Iranian Team Members (from left to right):

• Mina Dalirrooyfard

• Mohammad Pedramfar

• Erfan Tavakoli

• Mojtaba Shokrian Zini

• Sahand Seifnashri

• Alireza Fallah

This booklet is prepared by Meysam Aghighi and Masoud Shafaei

with special thanks to Morteza Sagha�an.

Copyright c©Young Scholars Club 2010-2011. All rights reserved.

1

28th

Iranian Mathematical Olympiad2010-2011

Contents

Problems 3First Round . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4Second Round . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5Third Round . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

Solutions 11First Round . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12Second Round . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14Third Round . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

2

Problems

3

First Round

1. (Morteza Adl) a and b are two integers such that a > b. If ab − 1 and a + b arerelatively prime, and so ab + 1 and a− b; prove that (ab + 1)2 + (a− b)2 is not aperfect square.

2. (Mohsen Jamali) There are n points in the plane that no three of them are collinear.Prove that the total number of triangles whose vertices are from these n pointsand having area equal to 1, is not greater than 2

3(n2 − n).

3. (Davood Vakili) Two circles W1 and W2 intersect at D and P . A and B are twopoints onW1 andW2 respectively such that AB is their common tangent. SupposeD is closer to AB than P . If AD intersectsW2 again in C, andM be the midpointof BC, prove that:

∠DPM = ∠BDC

4. (Arashk Hamidi) The coe�cients of P (x) = ax3 + bx2 + cx+ d are real and

min{d, b+ d} > max{|c|, |a+ c|}

prove that the equation P (x) = 0 does not have a solution in [−1, 1].

5. (Davood Vakili) In triangle ABC, ∠A = 60◦. Let E and F be points on theextensions of AB and AC such that BE = CF = BC. The circumcircle of ACEintersects EF in K (di�erent from E). prove that K lies on the bisector of ∠BAC.

6. (Mohammad Ali Karami) A school has n students andthere are some extra classes provided for them so thateach student can participate in any number of them.We know that there are at least two participants in anyclass. We also know that if two di�erent classes have twocommon students, then the number of their participantsare di�erent. Prove that the total number of classes isnot greater than (n− 1)2.

4

Second Round

1. (Mostafa Eynollahzadeh) Let P (x, y) be a polynomial in two variables with realcoe�cients. The degree of a monomial is sum of the powers of x and y in it. LetQ(x, y) be sum of the expressions in P (x, y) with highest degree. (For example ifP (x, y) = 3x4y − 2x2y3 + 5xy2 + x − 5 then Q(x, y) = 3x4y − 2x2y3.) Supposethere exists real numbers x1, x2, y1 and y2 such that

Q(x1, y1) > 0 , Q(x2, y2) < 0

prove that the set {(x, y)|P (x, y) = 0} is not bounded. (A subset S of the plane isbounded if there exists a positive numberM such that the distance of the elementsof S from origin is less than M .)

2. (Sina Jalali) We have put a rectangular parallelepiped with dimensions (2a+ 1)×(2b+ 1)× (2c+ 1) on an in�nite lattice plane with unit squares, where the verticesof its bottom face are lattice points, and a, b and c are positive integers. We canroll the parallelepiped in every four directions, and its faces are divided into unitsquares while the square in the middle of each face is black colored. If this blacksquare lies on a square of lattice, then it would become black as well.Prove that if the numbers 2a + 1 , 2b + 1 and 2c + 1 are pairwise coprime, thenwe can make every square of the lattice black by rolling the parallelepiped.

3. (Sepehr Ghazi Nezami) A is a set of n points in the plane. A version of A is aset of points that can be reached from A by translation, rotation, homothety orany combination of these transformations. We want to place n versions of A inthe plane so that the intersection of every two of them be a single point, and theintersection of every three of them be empty.

(a) Prove that if no four points of A form a parallelogram, we can do this usingonly translations. (Note that A does not contain a parallelogram with zeroangle, or a parallelogram that its opposite vertices coincide.)

(b) Prove that we can always do this using above transformations.

5

4. (Sepehr Ghazi Nezami) Suppose S be a �gure in theplane that its border does not contain any lattice pointand let x, y ∈ S be two lattice points with unit distancefrom one another. A copy of S is a combination of ro-tations and re�ections of S. Assume that there existsa tiling of the plane with copies of S such that x andy lie on lattice points for every copy of S in this tiling.Prove that the area of S is equal to the number of latticepoints in it.

5. (M.H.Sha�nia, M.Mansouri, S.Dashmiz) Let n be a positive integer and x1, x2, ...a sequence of 1 and -1 with these properties:

• is periodic with 2n − 1 as its fundamental period. (i.e. for every positiveinteger j we have xj+2n−1 = xj and 2n − 1 is the minimum number with thisproperty.)

• there exist distinct integers 0 ≤ t1 < t2 < ... < tk < n such that for every jwe have

xj+n = xj+t1 × xj+t2 × ...× xj+tk

prove that for every positive integer s < 2n − 1 we have

2n−1∑i=1

xixi+s = −1

6. (Sepehr Ghazi Nezami) We de�ne a 12-gon in the plane to be good if it has theseconditions:• be regular.

• be �lled.

• the origin be its center.

• its vertices contain points (0, 1) ,(1, 0) , (−1, 0) and (0,−1).

Find the number of faces of the polyhe-dral with maximum volume that its pro-jections on xy , yz and zx planes are good12-gons. (It is clear that the centers ofthese 12-gons coincide and is the origin ofthe 3D-space.)

6

7. (Bijan Aghdasi) S is a set with n elements, and P (S) is the set of all subsets ofS. Let

f : P (S) −→ N

be a function with these characteristics:

• For each A ⊆ S we have f(A) = f(S − A).

• For every two A,B ⊆ S we have max(f(A), f(B)) ≥ f(A ∪B).

prove that the number of elements in the range of f is not greater than n.

8. (Mohsen Jamali) Show that there exist in�nitely many positive integers n suchthat n2 + 1 has no divisor of the form k2 + 1 except for n2 + 1.

7

Third Round

1. (Hesam Rajabzadeh) In non-isosceles triangle ABC, let M be the midpoint ofBC, and let D and E be the foot of perpendiculars from C and B to AB and ACrespectively. If L and K be the midpoints of MD and ME, and T be a point onLK such that AT ||BC, prove that TA = TM .

2. (Morteza Sagha�an) For what integer values of n > 2, there exist numbers a1, a2,..., an ∈ N such that the following sequence is a nonconstant arithmetic progres-sion?

a1a2, a2a3, ..., ana1

3. (Mohsen Jamali) There are n points on a circle (n > 1). We de�ne interval to bean arc between any two of these n points. Let F be a family of these intervalssuch that every element of F is a subset of at most one other element of F . Aninterval is called maximal if it is not a subset of any other interval. Let m denotethe number of maximal elements of F , and a be the number of its non-maximalelements, prove that:

m+a

2≤ n

4. (Mohammad Mansouri) We de�ne a �nite subset of N like A to be good if it hasthe following properties:

• For every three distinct a, b, c ∈ A we have gcd(a, b, c) = 1.

• For every two distinct b, c ∈ A, there exists a ∈ A that a 6= b, c and a|bc.

�nd all good sets.

5. (Mohammad Jafari) Find all surjective functions f : R −→ R such that for everyx, y ∈ R we have:

f(x+ f(x) + 2f(y)) = f(2x) + f(2y)

6. (Sepehr Ghazi Nezami) Let OBC be an isosceles triangle (OB = OC). Considercircle ω with center O and radius OB. Tangents to this circle from B and Cintersect at A. Consider circle ω1 inside triangle ABC which is tangent to ω, andtangent to AC at H. Circle ω2 is also inside triangle ABC and is tangent to ω,tangent to ω1 at J , and tangent to AB at K. Prove that the bisector of ∠KJHpasses through I, the incenter of triangle OBC.

7. (Ali Khezeli) Find the locus of all points P inside an equilateral triangle havingthe following property √

h1 +√h2 =

√h3

where h1, h2, h3 are the distances of P from sides of the triangle.

8

8. (Mahyar Se�dgaran) p is a prime number and k ≤ p. f(x) is a polynomial withinteger coe�cients that for every integer x, f(x) is divisible by pk. Prove thatthere exist polynomials A0(x), A1(x), ..., Ak(x) with integer coe�cients such that

f(x) =k∑

i=0

(xp − x)ipk−iAi(x)

also prove that the statement is not correct for k > p.

9. (Morteza Sagha�an) There are n points in the plane,not all on the same line. A line l in the plane is calledwise if one can divide these n points into two sets A andB (A ∩ B = ∅) such that sum of distances of points ofA from l is equal to the sum of distances of points of Bfrom l. Prove that there are in�nite points in the planesuch that n+ 1 wise lines pass through them.

10. (Mohammad Jafari) Find the smallest real k such that for every a, b, c, d ∈ R wehave: √

(a2 + 1)(b2 + 1)(c2 + 1) +√

(b2 + 1)(c2 + 1)(d2 + 1)

+√

(c2 + 1)(d2 + 1)(a2 + 1) +√

(d2 + 1)(a2 + 1)(b2 + 1)

≥ 2(ab+ bc+ cd+ da+ ac+ bd)− k

11. (Sepehr Ghazi Nezami) Let A′, B′ and C ′ denote the midpoints of sides BC, ACand AB of triangle ABC respectively. Assume that P and P ′ are two variablepoints such that PA = P ′A′, PB = P ′B′ and PC = P ′C ′. Prove that the linePP ′ passes through a �xed point.

12. (Mohsen Jamali) Let f : N −→ N be a function such that for every a, b ∈ N,af(a) + bf(b) + 2ab is a perfect square. Prove that for every a ∈ N we must havef(a) = a.

9

10

Solutions

11

First Round

1. We know that

(a+ b)2 + (ab− 1)2 = a2b2 + a2 + b2 + 1 =⇒ c2 = (a2 + 1)(b2 + 1)

now we prove that (a2 + 1, b2 + 1) = 1. Suppose the contrary, hence there exists aprime number p such that:

p|a2 + 1, p|b2 + 1 =⇒ p|a2 − b2

=⇒ p|a− b or p|a+ b

If p|a− b, then p|ab− b2 and since p|b2 + 1 then p|ab+ 1 which is a contradictionas we know (ab+ 1, a− b) = 1. The case p|a+ b similarly reaches contradiction.Now that a2 +1 and b2 +1 are relatively prime and their multiplication is a perfectsquare, both should be perfect squares which is impossible because a2 + 1 can notbe one and the proof is complete.

2. lemma 1: Let points A and B in the plane (with distance d), if there is a pointC such that the area of triangle ABC is 1, then the distance of C from line AB is2d. i.e. the locus of such points consist of two lines parallel to AB at a 2

ddistance

from it.proof: trivial.lemma 2: The number of triangles with vertices A and B having area 1 is atmost 4.proof: If there are more than 4 triangles with this property, then from lemma 1and pigeonhole principle it concludes that 3 of the points must lie on one of thetwo lines at distance 2

dfrom AB which is a contradiction.

Now we count the number of triangles (T ) in this way that for each pair of pointsthere can be at most 4 triangles and we have counted each point at least 3 times,hence

T ≤(n2

)× 4

3=⇒ T ≤ 2

3× n(n− 1)

3. Let the extension of PD intersect AB at N . Since ∠BDC = ∠BPC, it is enoughto prove that ∠DPM = ∠BPC or equivalently ∠DPB = ∠MPC. N is themidpoint of AB (because NA2 = ND ×NP = NB2).Now we have

∠PBC = ∠PDC =

_

AP

2= ∠PAB

∠PCB =

_

PB

2= ∠PBA

hence 4PCB and 4PBA are similar, and since PM and PN are their medians,the angles ∠DPB = ∠NPB = ∠MPC.

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4. Suppose that α be a real root for P (x) such that |α| ≤ 1, hence

P (α) = 0 =⇒ aα3 + bα2 + cα + d = 0 =⇒ aα3 + cα = −(bα2 + d)

=⇒ |α||aα2 + c| = |bα2 + d| =⇒ |aα2 + c| ≥ |bα2 + d| (∗)

on the other hand f(x) = |ax+ c| takes its maximum over the interval [0, 1] in oneof its boundary points, i.e. f(x) ≤ max{|c|, |a+ c|}, so

|aα2 + c| ≤ max{|c|, |a+ c|}

it follows from the hypothesis that d and b + d are positive, so bx + d > 0 foreach 0 ≤ x ≤ 1, hence similarly we can say that (considering minimum of g(x) =|bx+ d|)

|bα2 + d| ≥ min{|d|, |b+ d|}

from these inequalities and (∗) we have

max{|c|, |a+ c|} ≥ |aα2 + c| ≥ |bα2 + d| ≥ min{|d|, |b+ d|} ≥ min{d, b+ d}

hencemax{|c|, |a+ c|} ≥ min{d, b+ d}

and this is a contradiction, so there is no α with such property.

5. Let T be the intersection of BF and CE. We have ∠BCE + ∠BEC = ∠ABC,since BC = BE we have ∠BCE = ∠ABC

2. similarly ∠CBF = ∠ACB

2. Hence

∠CTF = ∠BCE + ∠CBF =∠ABC + ∠ACB

2= 60◦

and the quadrilateral ABTC is cyclic, so ∠EBF = ∠ACE = ∠AKE and then∠ABF = 180− ∠EBF = 180− ∠AKE = ∠AKF , thus ABKF is cyclic.Having ∠EBK = ∠CFK and ∠BEK = ∠FCK, and from BE = FC we con-

clude that 4KBE and 4KFC are congruent, so KC = KE =⇒_

KC =_

KE,thus AK is the bisector of ∠BAC.

6. Suppose Ai (2 ≤ i ≤ n) be the set of all classes with i participants. We will

show that |Ai| ≤ n(n−1)i(i−1) . Since every 2-element subset of students are together in

at most one class of each Ai, then the number of 2-element subsets of Ai multiplythe elements of Ai must be less than or equal to the number of 2-element subsets

of students, hence |Ai| ≤(n2)

(i2)

= n(n−1)i(i−1) . On the other hand m = |A2| + ... + |An|,

hence we have

m ≤ n(n− 1)(1

2(2− 1)+ ...+

1

n(n− 1)) = n(n− 1)(1− 1

n) = (n− 1)2

13

Second Round

1. First note some points: (in the proof by the ray passing through origin and (x, y),we mean the ray with origin as its start point)

• We can suppose that (0, 0) , (x1, y1) and (x2, y2) are not collinear, otherwisechoose another point (x, y) out of this line such that Q(x, y) 6= 0 (if suchpoint doesn't exist then Q must be zero on all plane except a line, hence itmust be zero overall.) and with respect to the fact that the sign of Q(x, y)is di�erent from the sign of either Q(x1, y1) or Q(x2, y2), let it be Q(x1, y1),now we can replace (x2, y2) by (x, y).

• We know that Q is homogeneous so if we put k = deg(Q) we have Q(tx, ty) =tkQ(x, y) and we have tk > 0 for t > 0 so Q(tx, ty) and Q(x, y) have the samesign for t > 0.

• If two rays passing through origin doesn't form a line together, then thereexists point (s1, s2) on the �rst ray, and point (t1, t2) on the second ray suchthat the distance of origin from the line passing through these points is greaterthan an arbitrary M ∈ R+. (this is trivial.)

lemma. For every (x, y) ∈ R2 that Q(x, y) 6= (0, 0), there exists S ∈ R+ such thatif A > S then P (Ax,Ay) and Q(x, y) are of the same sign.proof of lemma. Note that if degQ = k then there exist homogeneous polyno-mials Q0, ..., Qk−2, Qk−1 such that degQi = i (0 ≤ i ≤ k − 1) and

P = Q+Qk−1 +Qk−2 + ...+Q0

let L(A) = P (Ax,Ay) , F (A) = Q(Ax,Ay) and Fi(A) = Qi(Ax,Ay), now since Qand Qi are homogeneous we have

L(A) = P (Ax,Ay) = AkQ(x, y) + Ak−1Qk−1(x, y) + ...+Q0(x, y)

that Qi(x, y) (1 ≤ i ≤ k−1) and Q(x, y) are constant numbers, so when A→ +∞the sign of L(A) is the sign of the coe�cient of Ak which is Q(x, y). (we assumedthat Q(x, y) 6= 0) So the lemma is proved.

Now with respect to above notes there is a point on the ray passing through originand (x1, y1) like (l1, l2), and another point on the ray passing through origin and(x2, y2) like (m1,m2) that P (l1, l2) > 0, P (m1,m2) < 0, and the distance of originfrom the line passing through these points is greater than an arbitrary M ∈ R+.Now since the restriction of P to this line is a single variable polynomial (byreplacing x = ay + b or y = ax + b) that is negative in a point and positive inanother, so P has a root on this line and the problem is solved.

14

2. Suppose that at the beginning the cuboid lies on theplane like the �gure. By rolling it in the directions down,right, up, left, up, right respectively, it is translated bythe vector (2a+1, 2a+1) and lies on the same face. Nowwith respect to the symmetry between a and c, we cantranslate it by the vector (2c + 1, 2c + 1), lying on thesame face as well. Now we show that this can also beachieved for b by rolling it upward, translating by thevector (2b+ 1, 2b+ 1), then rolling it downward.

Now since the center of the cuboid face is of integer values, if we can color a cellof the lattice plane with (x, y) coordinates, we can color these cells as well

(x+ 2a+ 1, y + 2a+ 1)

(x+ 2b+ 1, y + 2b+ 1)

(x+ 2c+ 1, y + 2c+ 1)

now we prove that we can color (x+1, y+1) too. As 2a+1 and 2c+1 are coprime,there exists r, s ∈ Z such that r(2a+ 1) + s(2c+ 1) = 1, now we can write

(x+ 1, y + 1) = (x+ r(2a+ 1) + s(2c+ 1), y + r(2a+ 1) + s(2c+ 1))

so this cell can be colored. If the center of the bottom face of the cuboid lies onthe (0, 0) cell of the plane, we can color all (x, y) cells where x+y is even. Withoutloss of generality assume that a and b have the same parity and a + b is even, soa + b + 1 is odd. Now by rolling the cuboid in the shown direction we can colora cell that sum of its coordinates are di�erent from the previous cell, and withrespect to what is said, we can color all of the plane cells.Remark. The statement also holds for the hypothesis (2a+ 1, 2b+ 1, 2c+ 1) = 1.

3. (a) Let an arbitrary point of the plane be origin, and suppose that the pointsof the set A be {a1, a2, ..., an}. We de�ne S + b = {x + b|x ∈ S} whereS ⊆ R2 , b ∈ R2. Now we prove that the sets A + a1 , A + a2 , ... andA + an have this property. First note that these are n translations of A.Let Bi = A + ai (1 ≤ i ≤ n), since aj + ai ∈ Bi and ai + aj ∈ Bj, soxij = ai +aj ∈ Bi∩Bj. Now if x ∈ Bi∩Bj and x 6= xij, then there must existj1 6= j and i1 6= i that ai1 +aj = x = aj1 +ai, hence x−xij = aj1−aj = ai1−ai,so ai,ai1 ,aj and aj1 form a parallelogram, a contradiction, therefore Bi ∩ Bj

has a single element.Clearly, with respect to previous part, the intersection of every three of themis empty, because if x ∈ Bi1∩Bi2∩Bi3 , then we must have ai1 +ai2 = ai1 +ai3 ,so ai2 = ai3 and i2 = i3 a contradiction. Therefore part (a) is complete.

(b) Let an arbitrary point apart from a1, ..., an be origin. Assume that a1, ..., anare complex numbers in this plane. Similar to the previous part we de�ne

15

S.b = {xb|x ∈ S} where b ∈ C and S ⊆ C is arbitrary. Let Bi = A.ai , (1 ≤i ≤ n), similar to part (a) we have

ai.aj = aj.ai ∈ Bi ∩Bj

now for every i1,i2,i3 and i4 that at least three of them are distinct we musthave ai1ai2 6= ai3ai4 , otherwise

ai1ai3

=ai4ai2

and it means that the origin is

the center of a rotational homothety that brings the segment ai1ai3 to thesegment ai2ai4 . Clearly since the ways of choosing i1,i2,i3 and i4 are �nite, wecan choose O in such a way that this won't happen, and the Bi ∩ Bj wouldhave one element in this case. Note that each one of the B1, ..., Bn is a versionof S, since it is obtained from S by a rotational homothety about the origin.(note that it is required that no two of ais are collinear with origin for takingit as center of this rotational homothety)

4. As said in the hypothesis, there are two points x, y with unit distance on a tile thattheir corresponding points on other tiles are lattice points. Let z be another latticepoint in S, and let S ′ be a copy of S. Also let x′, y′, z′ ∈ S ′ be the correspondingpoints to x, y, z. There are four possible places for z′ shown in below �gure, andall of them are lattice points, therefore the total number of lattice points of eachof the copies of S are equal, let this number be k.

Let d ∈ N be greater than the maximum distance between points of S. Considerthe following squares of size t× t and (t+ 2d)× (t+ 2d).

Assume that u be the number of tiles that has intersection with t× t square, sinceevery tile has at most k lattice points of that square we have u ≥ (t+1)2

k, on the

other hand none of the tiles has any point outside the (t+ 2d)× (t+ 2d) square, so

16

we have u ≤ (t+2d+1)2

k. Now let s be the area of S, since the inner square is totally

covered we have us ≥ t2 and us ≤ (t+ 2d)2, hence

(t+ 1)2 ≤ uk ≤ (t+ 2d+ 1)2

t2 ≤ us ≤ (t+ 2d)2

}⇒ (t+ 1)2

(t+ 2d)2≤ k

s≤ (t+ 2d+ 1)2

t2

⇒ limt→∞

(t+ 1)2

(t+ 2d)2≤ k

s≤ lim

t→∞

(t+ 2d+ 1)2

t2⇒ k = s

so the proof is complete.

5. We de�ne [x]i = xi where x = (x1, · · · , xn).Lemma. Let n, k, 0 < i1 < i2 < · · · < ik ≤ n be integers, then we have∑x∈X′

[x]i1 [x]i2 · · · [x]ik = −1 where X ′ = X − {(1, 1, · · · , 1)︸ ︷︷ ︸n

} and X is the set of all

n-tuples of {−1, 1}.Proof of lemma. Since for every n-tuple x, there exists a n-tuple y such that

[x]i1 6= [y]i1 and [x]s = [y]s for all s 6= i1, then we have∑x∈X

[x]i1 [x]i2 · · · [x]ik = 0.

This proves the lemma.Because 2n− 1 is the minimum period of the sequence, the n-tuples (xi, xi+1, · · · ,xi+n−1) for 1 ≤ i ≤ 2n − 1 are distinct. ((1, 1, · · · , 1)︸ ︷︷ ︸

n

is not among them) Note

that by de�ning xi = x2n−1+i we can expand the sequence to negative indices.Also note that xixi+s can be written in the form of product of xi−1, · · · , xi−n, be-cause from the hypothesis we have xi = xi−n+t1 × · · · × xi−n+tk , xi can be writtenin the form of product of xi−1, · · · , xi−n, now xi+s can be written in the form ofproduct of xi+s−1, · · · , xi+s−n, xi+s−1 can be written in the form of product ofxi+s−2, · · · , xi+s−n−1 and so xi+s can be written in the form of xi−1, · · · , xi−n. In-deed every member of the sequence can be evaluated using n consecutive membersof this sequence, now since x2l = 1 for every l ∈ Z, according to what we said,there exist 0 < u1 < u2 < · · · < up ≤ n such that for every l ∈ Z we have

xp = xl−u1 × xl−u2 × · · · × xl−up

hence2n−1∑i=1

xixi+2 is in the form of one of the summations stated in the lemma,

because i can be changed from n+ 1 to n+ 1 + 2n− 2 and xixi+s can be evaluatedby n members before xi, and these n members form a n-tuple and we know that bychanging i these sequences will contain all n-tuples of {−1, 1} except (1, 1, · · · , 1)︸ ︷︷ ︸

n

so using lemma, this summation is equal to -1.

6. We show that the desired polyhedron has 36 faces. It is trivial that this polyhedronis in the intersection of three 12-gonal prisms that their projections on the axisplanes are mentioned 12-gonals. Now for constructing the shape with maximum

17

volume we should consider the intersection of these three prisms. First considerthe prism corresponding to xy plane, the resulting shape is a 12-gon, now considerthe intersection of this shape with the prism corresponding to xz plane, we willshow that each face of the recent prism, adds exactly one face to the shape andthe intersection of this two prisms forms a polyhedron with 24 faces. For showingthis it is su�cient to show that each face of the xz prism, intersects the xy prism,because the xy prism and every face are convex, the desired intersection is alsoconvex, so it is connected and only adds one face. (because the projection of xyprism on xz plane is a �nite strip, it contains the 12-gon on xz plane)So the intersection of these two prisms is a polyhedron with 24 faces which is con-vex, because it is the intersection of two convex shapes. On the other hand thispolyhedron contains the points (x, y, z) = (0, 1, 1), (0, 1,−1), (0,−1, 1), (0,−1,−1),so it contains a square with vertices at these points, and the projection of this poly-hedron on the yz plane totally contains the 12-gon corresponding to yz plane.Similarly to the previous part, if we consider the intersection of this polyhedronand the yz prism, �rstly, every face of this prism, intersects the polyhedron, sec-ondly since the polyhedron is convex, every face intersects the polyhedron in oneregion and adds one face to the shape.So for every face of the 12-gonal prism, one face adds to the polyhedron, thereforethe resulting shape has 24 + 12 = 36 faces.

7. lemma. If B is obtained by intersection and complement from A1, A2, ..., Am,then

f(B) ≤ max{f(A1), f(A2), ..., f(Am)}

proof. It is su�cient to prove that f(A ∩B) ≤ max{f(A), f(B)},

f(A ∩B) = f((A ∪B)c) = f(Ac ∩Bc)

≤ max{f(Ac), f(Bc)} = max{f(A), f(B)}

and the proof is complete.First Solution. Suppose that f(A1) < f(A2) < ... < f(Am), using lemma, itconcludes that Am is not obtained from A1,A2,...,Am−1. (by intersection and com-plement) Since Am is obtained from single-element sets of itself, so there existsi ∈ Am such that {i} is not obtained from A1,A2,...,Am−1. By replacing some of Ajs(1 ≤ j ≤ m−1) with their complement we can assume that i ∈ A1∪A2∪...∪Am−1,and since i is not obtained from A1,A2,...,Am−1, so {i} 6= A1∪A2∪ ...∪Am, there-fore there exists l 6= i such that l ∈ A1 ∪ A2 ∪ ... ∪ Am. Now let a = {i, l}(indeed, construct a new element by joining i and l), if we restrict f to thesubsets of S that either have both i and l or none of them, then the functionf ′ : P ((S − {i, l}) ∪ {a}) −→ N is obtained and clearly has the hypothesis prop-erties. Hence, because A1, A2, ..., Am−1 ∈ Domain(f), from the hypothesis ofinduction we have m− 1 ≤ n− 1 so m ≤ n and the proof is complete.Second Solution. It concludes from the lemma that f(A∆B) ≤ max{f(A), f(B)},

18

now note that P (S) with operation ∆ forms a group (∅ as identity and A itself asits inverse), with respect to this the sets

Bi = {x|f(x) ≤ i} (1 ≤ i ≤ m)

are subgroups of P (S) (let {1, 2, ..,m} be the range of f). Therefore by Lagrange'sTheorem |Bi|||P (S)|, so |Bi| is a power of 2, and since B1 ( B2 ( ... ( Bm = P (S)and B1 has at least two elements (x ∈ B1 ⇒ xc ∈ B1) hence

|B1| ≥ 21, |B2| ≥ 22, ..., |Bm| ≥ 2m =⇒ 2n ≥ 2m =⇒ m ≤ n

therefore the proof is complete.Note that there is a f that its range has exactly n elements, let

f(A) := max{min(A),min(Ac)}

and by min{X} we mean the minimum element of X.

8. We claim that for every positive integer n, n2 + 1 has a divisor of the form k2 + 1that satis�es the statement of the problem. If n2 + 1 has no divisor of the formk2 + 1 (except n2 + 1) we are done. Otherwise suppose that k2 + 1|n2 + 1 for somepositive integer k. If k2 + 1 has no divisor of this form we are done. Otherwisel2 + 1|k2 + 1, by continuing this way our claim is proved.Now it is su�cient to consider an in�nite sequence an of positive integers of theform k2 + 1 such that (am, an) = 1 whenever m 6= n. As an example consideran = 22n + 1.

19

Third Round

1. Let H be the orthocenter of ABC, and let ω be the circle with diameter AH. SinceTA ⊥ AH, TA is tangent to ω in A. On other hand BM = EM ⇒ ∠MEB =

∠MBE = 90−∠C = ∠HAC =_EH2

(of ω), hence ME is tangent to ω and so isMD. Thus LK is the radical axis of point M and circle ω, and so the powers ofT with respect to M and ω are equal, which means

TA2 = TM2 =⇒ TA = TM

2. Let n be even, in this case no such numbers exist because if n = 2k we havea1a2 < a2a3 < ... < a2k−1a2k < a2ka1 hence a1 < a3 < ... < a2k−1 < a1 which is acontradiction.Now let n be odd, we construct such progression with a1, ..., a2k+1 ∈ R+, assumeb1 = x and

b1b2 = 1, b2b3 = 2, ..., b2k+1b1 = 2k + 1

b1b2 = 1⇒ b2 =1

x⇒ b3 = 2x⇒ b4 =

3

2x, · · ·

b2i = 1×3×···×(2i−1)2×4×···×(2i−2) ×

1x

2 ≤ i ≤ k

b2i+1 = 2×4×···×(2i)1×3×···×(2i−1)x 1 ≤ i ≤ k

but b2k+1b1 = 2k + 1 so

2× 4× · · · × (2k)

1× 3× · · · × (2k − 1)x2 = 2k + 1⇒ x =

√1× 3× · · · × (2k + 1)

2× 4× · · · × (2k)

now let ai = bi√

(2k + 1)!, thus it is trivial that a1a2, a2a3, · · · , ana1 forms anarithmetic progression too. On the other hand

a2i+1 =2× 4× · · · × (2i)

1× 3× · · · × (2i− 1)×x×

√(2k + 1)! =

(2×· · ·×(2i)

)((2i+1)×· · ·×(2k+1)

)a2i =

1× 3× · · · × (2i− 1)

2× 4× · · · × (2i− 2)× 1

x×√

(2k + 1)! =(1×3×· · ·×(2i−1)

)((2i)×· · ·×(2k)

)therefore all of ai's are distinct natural numbers, and the statement is true for alln = 2k + 1.

3. Let the points be 1, 2, ..., n. We denote an interval by [i, j] where i, j ∈ {1, 2, ..., n}and i < j. Let A = {[x1, y1], [x2, y2], ..., [xm, ym]} be the set of maximal intervals ofF , we claim that ∀1 ≤ i, j ≤ m : xi 6= xj. If xi = xj then one of their correspondingintervals must contain the other one, so the claim is proved. Similarly ∀1 ≤ i, j ≤m : yi 6= yj. Let X, Y denote the multi-set (the members of a multi-set need notto be distinct) of all lower and upper endpoints of all intervals of F respectively,and let [x, y] be a non-maximal interval of F , we show that either x /∈ X − {x}

20

or y /∈ Y − {y}. If x ∈ X − {x} and y ∈ Y − {y}, there must exist intervals [x, a]and [b, y] in F di�erent from [x, y]. Let

B = [x, y], C = [x, a], D = [b, y]

if B ⊆ C,B ⊆ D then two intervals contain B, a contradiction. If B ⊆ C,B * D,then we must have D ⊆ B and so D ⊆ B,C, hence two intervals contain D, acontradiction. Similarly B * C,B ⊆ D is not possible. And if B * D,B * Cwhich means D ⊆ B,C ⊆ B, therefore C,D are not maximal and since B is notmaximal either there exists E ∈ A such that D ⊆ B ⊆ E, so D is subset of twointervals, a contradiction. Now let A1 denote the set of non-maximal intervals thattheir upper endpoints are di�erent from the upper endpoints of other elements ofF , and let A2 denote the set of non-maximal intervals that their lower endpointsare di�erent from the lower endpoints of other elements of F , so |A1| + |A2| ≥ a.This results that the upper endpoints of intervals in A and A1 are distinct, so wehave |A| + |A1| ≤ n, hence m + |A1| ≤ n. Similarly m + |A2| ≤ n, and since|A1|+ |A2| ≥ a, either |A1| ≥ a

2or |A2| ≥ a

2, so m+ a

2≤ n.

4. First note that 1 /∈ A because if 1 ∈ A and a be the minimum element of A−{1},then there must exists b ∈ A, b 6= 1, a that b|a and this means that 1 < b < awhich is a contradiction.Now we prove that if a, b ∈ A and |A| ≥ 4, then a - b. Assume that a|b, then forevery c ∈ A− {a, b} we must have

(a, c) = (a, b, c) = 1

now there exists d ∈ A such that d|ac, we claim that b|ac, if b - ac, since d ∈A−{a, b} and (d, a) = 1, hence d|c. Similarly there exists e ∈ A that e|ad, if e = bthen

b|ad|acwhich is contradiction, so our claim is proved, therefore e ∈ A − {a, b} so e|d,hence if e = c then c = d which can't be true, otherwise e|d|c which means(e, d, c) = e > 1, a contradiction (note that the existence of e needs A to haveat least 4 elements, for |A| = 4 there isn't such e and we lead to contradictionsooner) Thus b|ac so if c and d be two elements of A− {a, b} we have

b|acb|adb|ab

⇒ b|a(b, c, d) = a⇒ b|a⇒ a = b

which is a contradiction. So assume that |A| ≥ 4, in this case a ∈ A exists suchthat for two pairs {b, c} and {d, e} we have

a|bc, a|de

because we can choose these pairs in(n2

)ways, and if n ≥ 4 we have

(n2

)> n. Now

if b, c, d, e be distinct, and let p be a prime factor of a, hence p|d or p|e, suppose

21

p|d, similarly suppose that p|c, therefore p|(a, c, d) which is a contradiction, so e.gwe must have b = d, hence

a|bca|bea|ba

⇒ a|b(c, e, a) = b

which can't be true. Therefore |A| ≤ 3, now assume that A = {a, b, c} be good,so (a, b, c) = 1, a|bc, b|ac and c|ab, thus a = pq such that q|b and p|c. Let b = qrand c = pr′, we have

c|ab⇒ pr′|pq.qr ⇒ r′|q2rnote that if r′ and q have common factors, so do a, b, c unless q = 1 and it meansa|c, so (a, b) = (a, b, c) = 1, now because b|ac we have b|c which means that a, bare two coprime factors of c, yet c|ab so c = ab. Therefore one condition is thatc = ab and (a, b = 1, if q > 1 we must have (r′, q) = 1 so r′|r, similarly r|r′, sor = r′ which means that a = pq, b = qr and c = pr, but in this situation we musthave (p, q, r) = 1 because if a prime number s|(p, q, r) then s|(a, b, c). Thus we allsets are of the form

A = {a, b, ab} (a, b) = 1 OR A = {pq, qr, rp} (p, q, r) = 1

5. Since f(x) is surjective, there exists t ∈ R such that f(t) = 0, now let

x = y = t⇒ f(t) = 2f(2t)⇒ f(2t) = 0 (1)

x = 0⇒ f(f(0) + 2f(y)) = f(0) + f(2y)

x = t(1)⇒ f(t+ 2f(y)) = f(2y)

}⇒

f(f(0) + 2f(y))−f(t+ 2f(y))= f(0)

but 2f(y) takes all real values, so for every c ∈ R we have

f(f(0) + c)− f(t+ c) = f(0)

on the other hand

c = t(1)⇒ f(f(0) + t) = f(0) (2)

c = 0 ⇒ f(f(0)) = f(0) (3)

c = f(0)(2)⇒ f(2f(0)) = 2f(0) (4)

x = 0, y = 0 ⇒ f(3f(0)) = 2f(0) (5)

now by putting x = 0, y = f(0) in the �rst equation, we obtain from (3), (4) that

f(3f(0)) = 3f(0)(5)⇒ f(0) = 0 now if we put x = 0, it results that f(2f(y)) =

f(2y), so f(x+ f(x) + 2f(y)) = f(2x) + f(2f(y)). Since 2f(y) takes all values inR we can conclude that

f(x+ f(x) + y) = f(2x) + f(y) (6)

22

by putting y = 0 we have

f(x+ f(x)) = f(2x) (7)

also putting y = f(x+ f(x)) = f(2x) in (6) results

f(x+ f(x) + y) = f(x+ f(x) + f(x+ f(x))) = f(2x+ 2f(x))

= f(x+ f(x) + (x+ f(x))) = f(2x) + f(x+ f(x)) = 2f(2x)

on the other hand

f(x+ f(x) + y) = f(x+ f(x) + f(2x)) = f(2x) + f(f(2x))

so f(f(2x)) = f(2x), and since f(2x) takes all real values we have

∀a ∈ R : f(a) = a

6. Lemma. In the next �gure BB′ and CC ′ are diagonals of ω and E and F aremidpoints of arcs B′C and BC ′ respectively. N is a point on BE and P is a pointon CF such that K ′P and HN are perpendicular to BC. We prove that N lieson K ′B′ if and only if P lies on HC ′.

Proof of lemma. Using Menelaus' theorem in triangleBTQ and secant secant K ′B′:

BK ′

K ′Q.QB′

B′T.TN

NB= 1

but TNNB

= CHHS′ , so

BK ′

K ′Q.CH

HS=B′T

QB′

similarly the condition that P lies on C ′H, leads to

BK ′

K ′Q.CH

HS=C ′U

SC ′

and considering the identity B′TQB′ = C′U

SC′ the lemma isproved.

Now, draw a perpendicular from H to BC such that intersects BE at N andB′N intersects with circle at X and with AB at K ′. HC ′ intersects line CF at Pand circle at Y . According to above lemma, K ′P is perpendicular to BC. Also

23

∠CHN = ∠CXN = ∠CIN , because all are equal to half of arc B′C. So thepentagon XHCNI and similarly the pentagon BPIY K ′ are cyclic. Suppose thatJ ′ is the second intersection point of circumcircles of these pentagons. We have

∠J ′HA = ∠J ′IC = ∠J ′Y H

so circumcircle of J ′Y H is tangent to AC.On the other hand we know that if a circle istangent to AC at H and to ω at Y ′, then Y ′

is the inverse homothetic center of these twocircles and since C ′ and H are homotheticso Y ′ should be on C ′H, therefore Y = Y ′.Hence this circle is exactly the circumcircleof J ′Y H. So circumcircle of J ′Y H is tangentto ω. Similarly circumcircle of K ′XJ ′ is alsotangent to AB and ω. We have

∠K ′J ′H = ∠ACI+∠ABI = 180◦− ∠BAC2

= ∠BIC = ∠AHJ ′ + ∠AK ′J ′and since AH and AK ′ are tangent to circumcircles of J ′Y H and K ′J ′X, it fol-lows that circumcircles of J ′Y H and K ′J ′X are also tangent to each other. Socircumcircle of J ′Y H is ω1 and circumcircle of K ′J ′X is ω2. Also J ′ = J andK ′ = K and ∠IJH = 180◦ −∠ICH = 180◦ −∠IBK = ∠IJK, so IJ is bisector.

7. We claim that the answer is the incircle. We have

∠XDP + ∠XEQ =_DX2

+_XE2

= ∠DOE2

= 60◦

∠XFR − ∠XDP =_FD+

_DX

2−

_DX2

=_FD2

= ∠FOD =60◦ so if ∠XDP = α we have ∠XEQ = 60◦ − α and∠XFR = 60◦ + α. According to the law of sines

PX = XD sinα = (2r sinα) sinα = 2r sin2 αRX = XF sin∠XFR = XF sin(60 + α) = 2r sin2(60 + α)QX = XE sin∠XEQ = XE sin(60− α) = 2r sin2(60− α)

⇒√RX =

√2r sin(60 + α) =

√2r(sinα + sin(60− α)) =

√PX +

√QX

Now if this property holds for a point not on the incircle, without loss of generalitywe assume that the point is in the quadrilateral AEOD, let X ′ denote the point ofintersection of AX with the incircle (closer to A), a homothety with center A andAXAX′ as its coe�cient, mapsX ′, B, C toX,B′, C ′ respectively, according to previous

part since the incircle of A′B′C ′ passes throughX, we have√XR′ =

√XQ+

√XP ,

24

on the other hand we know that√XR =

√XQ +

√XP which su�ces that

XR′ = XR, hence X and X ′ coincide, which completes the proof.Remark. For an arbitrary triangle ABC, the desired locus is an ellipse tangentto sides of the triangle.

8. We prove the �rst part by induction ,for k = 1 the problem is well-known. Nowwe add the constraint that deg(Ai) ≤ p−1 for i ≤ k−1. Using division algorithmwe can write

f(x) = (xp − x)kAk(x) + g(x), deg(g) ≤ kp− 1

now the hypothesis of induction for k − 1 on g(x) results that

f(x) = (xp − x)kAk(x) + (xp − x)k−1Ak−1(x) + · · ·+ pk−1A0(x) (I)

⇒ p|(xp − xp

)k−1Ak−1(x) + (xp − xp

)k−2Ak−2(x) + · · ·+ A0(x)

if we let h(x) = xp−xp

, it is obvious that h has integer value but it does not haveinteger coe�cients. We have

g(x+ ap) ≡ (x+ ap)p − (x+ ap)

p≡ xp − x− ap

p≡ h(x)− a (mod p)

and Ai(x + ap) ≡ Ai(x) (mod p). Hence p|(h(x) − a)k−1Ak−1(x) + · · · , so for aconstant x, a takes all values module p, therefore

p|yk−1Ak−1(x) + · · ·+ A0(x) (II)

now let x be �xed, so we have a polynomial in terms of y whose degree is atmost k − 1 ≤ p− 1, thus its coe�cients are divisible by p, hence ∀x, i : Ai(x) ≡ 0(mod p). Now since the degree of Ai is less than or equal to p−1, we have Ai(x) =pBi(x) (III) in which the coe�cients of Bi(x) are integer. By replacement of (III)in (I) the statement results.For k > p the counterexample f(x) = (xp − x)k−1 − pp−1(xp − x)k−1 works.

9. lemma. Suppose that we have a set of n points is the plane and G be theircentroid, if a line l passes through G then sum of the signed distances of these npoints from l is zero. (i.e. sum of the distances of points in each side of the lineare equal)proof of lemma. Let G be the origin of the plane and l be the x-axis, since G isthe average of all of the points we have

0 = yG =yA1 + yA2 + ...+ yAn

n⇒ yA1 + yA2 + ...+ yAn = 0

so the proof is complete.Let A = {A1, A2, ..., An} be the points. Now we de�ne n + 1 points G,G1, ..., Gn

25

in this way: let G be the centroid of these n points and assume that O be anarbitrary point in the plane, let A′i be the re�ection of Ai over O, we de�ne Gi

to be the centroid of A1, ..., Ai−1, A′i, Ai+1, ..., An. According to what we said, the

line OG is a wise line for the point O, if we divide the points into two sets byOG. (we can assign the points on OG to each of the sets) We claim that OGi

is a wise line, if A′i be in one side of OGi with points {Ai1 , Ai2 , ..., Aik}, then letX = {Ai, Ai1 , Ai2 , ..., Aik} be one of the sets and Y = A − X be the other set,hence we have found n + 1 wise lines. Now we have to prove that there existsin�nite points like O that these lines are pairwise distinct for them.Suppose that for a point O, G and Gi are collinear with O, we have

~G =~A1+ ~A2+...+ ~An

n

~Gi =~A1+ ~A2+...+ ~An− ~Ai+ ~A′

i

n

}⇒ ~G− ~Gi =

~Ai − ~A′in

=2

n( ~Ai − ~O)

this means that O, Ai and G are collinear, thus it is su�cient that O does not lieon lines A1G,A2G, · · · , AnG. Therefore the assumption that O,G and Gi are notcollinear is correct.

Now, suppose that O,Gi and Gj are collinear, and−−→GGi = 2

n

−−→AiO. Hence the

triangles GGiGj and OAiAj are similar with ratio 2n, and assume that the center

of the homothety taking OAiAj to GGiGj is T . Now let the extension of AiAj

intersects OG in S, thus TGTO

= 2n, TO

TS= 2

n, therefore

GO

GS=GO/GT

GS/GT=

n/2 + 1

n2/4− 1=

1

n/2− 1=

2

n− 2

in this case a homothety with center G takes O to S and we have OGGS

= 2n−2 .

Therefore it is su�cient that O does not lie on the lines that are resulted byhomothety over G with ratio −2

n−2 from the lines AiAj.

10. According to Cauchy-Schwartz inequality we can write√(a2 + 1)(b2 + 1)(c2 + 1) =

√((a+ b)2 + (ab− 1)2)(c2 + 1)

≥ (a+ b)c+ ab− 1

⇒∑cyc

√(a2 + 1)(b2 + 1)(c2 + 1) ≥ 2

∑sym

ab− 4

and equality holds for a = b = c = d =√

3, so the minimum value for k is 4.

26

11. First Solution. Let the homothety with center at G, centroid of ABC andA′B′C ′, and ratio −1

2maps P to P1, then we have P1A′

A′P ′ = P1B′

B′P ′ = P1C′

C′P ′ = 12, so the

circumcircle of A′B′C ′ is the apollonius circle about P1 and P ′ with ratio 12(i.e.

the locus of the points X with the property that XP1

XP ′ = 12). Let N denote the

center of this circle, and let x = |P ′P1|, we have (S and T are intersection pointsof the apollonius circle with PP ′)

P1T = x3

P1S = x

}⇒ NS =

P1S + P1T

2=

2x

3⇒ NP1 =

x

3, NP ′ =

4x

3

note that NT = 2x3is the radius of the apollonius circle, so x is constant, hence the

locus of P1 and P′ are two circles with centers at N , on the other hand a homo-

thety with respect to G takes P to P1, so the locus of P is a circle centered at thecircumcenter of ABC. Now since P ′ maps to P1 with a homothety centered at Nhaving ratio 1

4, and P1 maps to P with a homothety with ratio −1

2, by combining

these two homotheties P ′ maps to P with a homothety with ratio −12, therefore

PP ′ passes through the center of this homothety which is a �xed point.

Second Solution. Let G be the origin, we have (so by ~K we mean−−→GK)

~A′ =− ~A2, ~B′ =

− ~B2, ~C ′ =

−~C2

the hypothesis results that

(~P − ~A)2 = ( ~P ′ − ~A′)2 ⇒ ~P 2 + ~A2 − 2 ~A. ~P = ~P ′2

+~A2

4+ ~A. ~P ′ (1)

(~P − ~B)2 = ( ~P ′ − ~B′)2 ⇒ ~P 2 + ~B2 − 2 ~B. ~P = ~P ′2

+~B2

4+ ~B. ~P ′ (2)

(~P − ~C)2 = ( ~P ′ − ~C ′)2 ⇒ ~P 2 + ~C2 − 2~C. ~P = ~P ′2

+~C2

4+ ~C. ~P ′ (3)

(1), (2) ⇒ 34( ~A2 − ~B2) = (2~P + ~P ′).

−→BA

(2), (3) ⇒ 34( ~B2 − ~C2) = (2~P + ~P ′).

−−→CB

note that the vectors−→BA and

−−→CB are independent vectors and the dot product of

them with 2~P + ~P ′ is a �xed value, so 2~P + ~P ′ is determined uniquely. Now let X

be a point on PP ′ such that XP ′

XP= 2, thus ~X = 2~P+ ~P ′

3, hence X is the �xed point

that PP ′ passes through.

12. First note that if p be an odd prime, putting a = b = p results that 2p(p+ f(p)) isa perfect square, so p|p + f(p) and p|f(p). Now suppose that for a ∈ N, af(a) isnot a perfect square, therefore for a prime p we have p2k−1‖af(a) that k ∈ N. Letaf(a) = p2k−1s where s ∈ N, p - s, now put b = p2k so p2k−1s+ p2k(f(p2k) + 2a) is

27

a perfect square, which means p2k−1(s+p(f(p2k) + 2a)) is a perfect square but theterm in the parenthesis does not have p factor which is a contradiction, so af(a)is a perfect square. Now assume that p > f(1) be an odd prime, since p|f(p) sop ≤ f(p), hence pf(p) + f(1) + 2p < (

√pf(p) + 2)2 ⇔ f(1) + 2p < 4

√pf(p) + 4,

because p ≤ f(p) so p ≤√pf(p), therefore 4

√pf(p) ≥ 4p > 2p + f(1). So we

have(√pf(p))2 < pf(p) + f(1) + 2p < (

√pf(p) + 2)2

so pf(p) + f(1) + 2p = (√pf(p) + 1)2, hence

f(1) + 2p = 2√pf(p) + 1

now since p|f(p) and pf(p) is a perfect square so f(p) = k2p that k ∈ N. Now ifk ≥ 2 we have

3p > f(1) + 2p = 2√pf(p) + 1 ≥ 4p+ 1

so f(p) = p for any odd prime p greater than f(1), but above equation simulta-neously proves that f(1) = 1, so for every odd prime p we have f(p) = p. Nowsuppose for a ∈ N we have f(a) 6= a, so (b = p in which p is and odd prime)(a+ p)2 6= af(a) + p2 + 2ap, hence [af(a) + p2 + 2ap]− (a+ p)2 = af(a)− a2, butthe absolute value of left side of the equation is greater than or equal to 2(a+p)−1because the di�erence of (a + p)2 with any perfect square except itself is at leastthis value, so for any odd prime we have |af(a) − a2| ≥ 2(a + p) − 1 which can'tbe true for a constant a, so f(a) = a.

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