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MT 430 Intro to Number Theory
PROBLEM SET 7
Due Thursday 4/18
√
Problem 1. Given that 18 = h4, 4, 8i, find the least positive solution
of x2 − 18y 2 = −1 (if any) and of x2 − 18y 2 = 1.
and 172 − 18 × 42 = 1. It
Solution: We have h4, 4i = 4 + 41 = 17
4
follows that (17, 4) is the least positive solution of x2 − 18y 2 = 1 and
x2 − 18y 2 = −1 has no solutions.
√
Problem 2. Given 29 = h5, 2, 1, 1, 2, 10i, find the least positive solution of x2 − 29y 2 = −1 (if any) and of x2 − 29y 2 = 1.
70
Solution: We have h5, 2, 1, 1, 2i = 13
and 702 −29×132 = −1. It follows
that (70,
solution √
of x2 − 29y 2 = −1.
√13) 2is the2 least possible
√ Since
2
(70 + 13 29) = 70 + 29 × 13 + 2 × 70 × 13 29 = 9801 + 1820 29, we
conclude that (9801, 1820) is the least positive solution of x2 − 29y 2 =
1.
Problem 3. Prove that there are infinitely many positive integers n
n(n + 1)
such that
is a perfect square.
2
Solution: Let (a, b) be a positive integer solution of the Pell’s equation
x2 −2y 2 = 1. Then a2 −2b2 = 1, and so for n = 2b2 we have n(n+1)/2 =
2b2 (2b2 + 1)/2 = a2 b2 .
Since the Pell’s equation x2 − 2y 2 = 1 has infinitely many solutions,
we are done.
Problem 4. Prove that there are infinitely many positive integers n
such that n2 + (n + 1)2 is a perfect square.
Solution: Note that n2 + (n + 1)2 = 2n2 + 2n + 1. Let (a, b) be a
positive integer solution of the Pell’s equation x2 − 2y 2 = −1. Clearly,
a is odd, so we write a = 2n + 1. Then (2n + 1)2 − 2b2 = −1 implies
that 2n2 + 2n + 1 = b2 , a perfect square! Since the Pell’s equation
x2 − 2y 2 = −1 has infinitely many solutions, we are done.
Problem 5. Prove that x2 −dy 2 = −1 has no integer solutions if d ≡ 3
(mod 4).
Solution: If d ≡ 3 (mod 4), then x2 − dy 2 ≡ x2 + y 2 (mod 4). The
only possibilities for x2 + y 2 (mod 4) are 0, 1, 2, which does not include
−1 ≡ 3 (mod 4).
1
2
PROBLEM SET 7
Problem 6 (Harder). Let d be a positive integer, which is not a perfect
square. Suppose N is such that x2 − dy 2 = N has one integer solution.
Prove that x2 − dy 2 = N has in fact infinitely many solutions.
Solution: Let (x0 , y0 ) be a solution of x2 − dy 2 = N . Then for every
solution (a, b) of the Pell’s equation x2 − dy 2 = 1, we have:
x20 − dy02 = N
a2 − db2 = 1
Multiplying these two equalities together we obtain:
(ax0 + dby0 )2 − d(bx0 + ay0 )2 = N.
It follows that every solution (a, b) of the Pell’s equation x2 − dy 2 = 1
gives rise to a solution (ax0 +by0 , bx0 +ay0 ) of the equation x2 −dy 2 = N .
Since the former has infinitely many solutions, we are done.
Definition. Integer solutions of the Diophantine equation
x2 + y 2 = z 2
are called Pythagorean triples. Note that if (x, y, z) is a Pythagorean
triple and an integer d divides any two of (x, y, z) then it divides all
three of them. Consequently, one is most interested in Pythagorean
triples (x, y, z) such that gcd(x, y) = gcd(x, z) = gcd(y, z) = 1. Such
triples are called primitive.
In the next two exercises, you will find all primitive Pythagorean
triples.
Problem 7. Suppose (x, y, z) is a primitive Pythagorean triple. Show
that exactly one of x and y is odd and the other is even.
Solution: We have to rule out the possibility that x and y have the
same parity. If x and y are both even, then z must be even. This
contradicts the assumption that (x, y, z) is primitive. If x and y are
both odd, then z must be even. Then x2 + y 2 = z 2 modulo 4 becomes
1 + 1 ≡ 0 (mod 4), which is absurd. We are done.
Problem 8. Suppose (x, y, z) is a primitive Pythagorean triple with x
even. Prove that z − y = 2s2 and z + y = 2t2 for some coprime integers
s and t.
Conclude that all primitive Pythagorean triples (x, y, z) are of the
form (x, y, z) = (2st, t2 − s2 , t2 + s2 ) for some coprime integers s and t.
PROBLEM SET 7
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Solution: We have x2 = z 2 − y 2 = (z − y)(z + y). It is easy to see that
gcd(z − y, z + y) = 2. Uniqueness of prime factorization now implies
that z − y = 2s2 and z + y = 2t2 for some coprime integers s and t.
Solving for z and y, we obtain y = t2 − s2 and z = t2 + s2 . From
x2 = (z − y)(z + y), we finally obtain x = 2st.