MATH 115 HW 7

DUE 8/4

1 Do problem 5.8 in your textbook.

Solution:One strategy is to find two perfect squares that can be written as a sum of twosquares and multiply them: for example, take 25 = 52 + 02 = 32 + 42 and132 = 132 + 02 = 122 + 52. Then

4225 = (5 · 13)2 + 02 = (5 · 12)2 + (5 · 5)2 = (13 · 4)2 + (13 · 3)2.

2 Do problem 5.9 in your textbook.

Solution:Suppose (p/q)2 + (s/t)2 = n. Clearing denominators, p2 + s2 = q2t2n. By thecondition proved in Section 5.7, q2t2n has the property that every prime ≡ 3 (mod4) has even exponent in q2t2n. Therefore the same is true of n. Hence n is a sumof two squares.

3 Do problem 5.10 in your textbook. (Hint: follow the method in the book asclosely as possible. Consider the congruence x2 + 2y2 ≡ 0 (mod p) to show that−2 must be a square mod p for there to be a solution to x2 + 2y2 = p. For theconverse, follow the method via continued fractions presented in the book.)

Solution:First suppose x2 +2y2 = p. This says that x and y are smaller than p, so p does notdivide x or y. We have x2 + 2y2 ≡ 0 (mod p), so dividing by y, we get (x/y)2 ≡ −2(mod p). Therefore the Legendre symbol

(−2p

)= 1. By using our standard facts

on calculating Legendre symbols, we see that this happens if and only if p ≡ 1 or 3(mod 8).

Conversely, suppose p ≡ 1 or 3 (mod 8). Then we have r ∈ Z such that r2 ≡ −2(mod p). By Lemma 5.7.5 in the book, there exist integers a, b such that 0 < b <

√p

and ∣∣∣∣−rp − a

b

∣∣∣∣ < 1

b√p.

Let c = rb + pa. Then

2b2 + c2 ≡ 2b2 + r2b2 ≡ (2 + r2)b2 ≡ 0 (mod p)

so 2b2 + c2 = kp for some positive integer k. Because b2 < p and c2 < p, wemust have 2b2 + c2 = p or 2b2 + c2 = 2p. If 2b2 + c2 = p we’re done. Otherwise

Date: Due 8/4.

1

2 DUE 8/4

2b2 + c2 = 2p, and so c = 2c′ is even. Then we have

2b2 + 4c′2 = 2p.

Dividing this equation by 2 we conclude that b2 + 2c′2 = p, and we are done onceagain.

4 Do problem 5.11 in your textbook. (Hint: can a number congruent to 3 (mod 4)be a sum of 2 squares?)

Solution:Of four consecutive numbers, one is congruent to 3 mod 4. Such a number cannotbe a sum of two squares, since the congruence

x2 + y2 ≡ 3 (mod 4)

has no solution (the only squares modulo 4 are 0 and 1).

5 Determine for which integers n the equation n = x2 + 2y2 has a solution inintegers x, y (use your result from Problem 3). (Hint: follow the method in thebook/presented in class, using the fact that x2 + 2y2 = (x +

√−2y)(x −

√−2y)

is the norm in the ring Z[√d], and hence is multiplicative to show that the set of

integers that are expressible as x2 + 2y2 is closed under multiplication.)

Solution:Consider the norm map N : Z[

√−2] → Z sending x +

√−2y to (x +

√−2y)(x −√

−2y) = x2 + 2y2. The image of this map is precisely the set of integers whichcan be written as x2 + 2y2 for integers x and y. Notice that this image is closedunder multiplication: this is because for a, b ∈ Z[

√−2], N(ab) = N(a)N(b) and so

the product of two elements which are norms is also a norm. Now we showed inProblem 3 that a prime is expressible as x2 + 2y2 if and only if p = 2 or p ≡ 1 or3 (mod 8). Also, for trivially the square of any prime is expressible in this form.Therefore the following condition on n guarantees that n = x2 + 2y2 for integers xand y:

• For every prime p ≡ 1 or 3 (mod 8), the exponent of p in n is even.

Conversely, suppose that n = x2 + 2y2, and take any prime p ≡ 1 or 3 (mod 8).Then x2 +2y2 ≡ 0 (mod p), and therefore by our work in problem 3, p | y and p | x.Therefore p2 | x2 + 2y2 = n. We can therefore write n/p2 = (x/p)2 + 2(y/p)2. Byinduction we may assume that the exponent of p in n/p2 is even, and therefore theexponent of p in n is even as well.

6 Find the two smallest positive solutions (x, y) to the equation x2 − 14y2 = 1.

Solution:We start computing partial convergents in the continued fraction expansion of

√14.

The first few are

3/1, 4/1, 11/3, 15/4, ...

Checking these one-by-one, we see that (x, y) = (15, 4) is the smallest positivesolution to x2 − 14y2 = 1. To get the next smallest positive solution, take the

MATH 115 HW 7 3

purely rational and purely irrational parts of (15 +√

144)2:

(15 +√

144)(15 +√

144) = 449 +√

14120.

Therefore the next smallest positive solution is (x, y) = (449, 120).

7 Let N be any non-zero integer. Prove that if x2−dy2 = N has one solution, thenit has infinitely many.

Solution:I forgot to mention that d is positive and square-free in this problem, and that Nis non-zero. Therefore x2 − dy2 = 1 has infinitely many solutions. Let (s1, t1) and(s2, t2) be two distinct solutions to x2 − dy2 = 1, and let (x, y) be a solution tox2 − dy2 = N . Then

(x +√dy)(s1 +

√dt1) = (xs1 + dyt1) +

√d(xt1 + ys1),

(x +√dy)(s2 +

√dt2) = (xs2 + dyt2) +

√d(xt2 + ys2)

are distinct solutions to x2 − dy2 = N . To see that they are distinct, one caneither use some linear algebra, or, if one is comfortable with fields, notice thatQ(√d) is a field and hence the equation (x + y

√d)((s1 − s2) +

√d(t1 − t2)) = 0

has no solution except s1 = s2, t1 = t2. Therefore there are infinitely many pairs(xs1 + dyt1, xt1 + ys1). Each one of these is a solution to x2 − dy2 = N , since

(xs1 + dyt1)2 − d(xt1 + ys1)2 = N((x +√dy)(s1 +

√dt1))

= N(x +√dy)N(s1 +

√dt1) = N · 1.

8 Show that if d ≡ 3 (mod 4), then the equation x2− dy2 = −1 has no solution forx and y integers.

Solution:Modulo 4, this equation says

x2 + y2 ≡ 3 (mod 4)

since d ≡ −1 (mod 4). This equation has no solutions for x and y, since the onlysquares mod 4 are 0 and 1.

9 Find a parametrization of the rational solutions of the equation x2 − 2y2 = 1.(Hint: start with the solution (x, y) = (1, 0).)

Solution:As discussed in class, every rational point (x, y) satisfying x2 − 2y2 = 1 lies on aunique line through (1, 0), and every such line contains at most one rational pointbesides (1, 0). So to parametrize rational points with x2 − 2y2 = 1, we should takethe general line y = m(x − 1) (we’ll deal with the vertical line x = 1 last) passingthrough (1, 0) with m ∈ Q, and see if it intersects the line x2−2y2 = 1 in a differentpoint. We have

x2 − 2m2(x− 1)2 − 1 = 0,

or simplifying a little bit,

(1− 2m2)x2 + 4m2x− (2m2 + 1) = 0.

4 DUE 8/4

Since 1− 2m2 = 0 has no solution in Q, this is always a quadratic, with solutionsgiven by

x = 1, x =−2− 4m2

2− 4m2.

as one finds by the quadratic formula followed by some simplification. So the

second x-value is given by (after changing signs) 2m2+12m2−1 . Plugging in the formula

y = m(x−1) and simplifying, we see that the line y = m(x−1) intersects the curvex2 − 2y2 = 1 in the unique second point

(x, y) = (2m2 + 1

2m2 − 1,

2m

2m2 − 1).

Finally consider the vertical line x = 1. Then we have the equation −2y2 = 0, ory2 = 0, so there is no second point of intersection of the vertical line x = 1 through(1, 0) with the curve x2 − 2y2 = 1. The formula above is a full parametrization ofthe rational points on x2 − 2y2 = 1 besides (1, 0).

10 Find a parametrization of the rational solutions of the equation y2 = x3 + 2x2.(Hint: start with the solution (x, y) = (0, 0).)

Solution:We again consider the points of intersection of the line y = mx with the curvey2 = x3 + 2x2. We get

m2x2 = x3 + 2x2,

or x3 + (2 −m2)x2 = 0. This has the solutions x = 0, x = m2 − 2. Notice thatthese are distinct points since 2 is not a rational square. This generates the secondpoint

(x, y) = (m2 − 2,m3 − 2m)

on the curve y2 = x3 + 2x2. Finally consider the vertical line x = 0. Then wemust have y2 = 0 be our equation, so we do not get a new point on the curvey2 = x3 + 2x2 by intersecting with the vertical line. The above formula is a fullparametrization of the rational points on y2 = x3 + 2x2 besides (0, 0).