MATH 313: SOLUTIONS HW4
Problem 1
Consider the equality x4 − y 4 = z 2 . Notice that it can be rewritten in the following form:
z 2 + (y 2 )2 = (x2 )2 .
By contradiction, assume that there exists a solution z0 , y02 , x20 . If such solution exists, we can
choose z0 , y02 , x20 to be pairwise coprime such that (z0 , y02 , x20 ) is a primitive Pythagorean triple. We
know that z0 and y02 have opposite parity.
CASE 1: Let z0 be odd and y02 be even. Then, we have the following equalities, where (p, q) = 1
and one is even and the other is odd:
y02 = 2pq
z0 = p2 − q 2
x20 = p2 + q 2 .
Note that we now have a new Pythagorean triple with the equality: x20 = p2 + q 2 . Hence, again,
we have the following inequalities where a, b have the same properties as p, q:
p = 2ab
q = a2 − b 2
x 0 = a2 + b 2 .
We can use substitution to solve for y02 :
y0 2
) = ab(a2 − b2 )
2
But since (a, b) = 1, we have a, b, and a2 − b2 are pairwise coprime and thus each term has to be
02
2
2
02
0
0
a perfect square. Hence, we can rewrite: a = x02
0 , b = y0 and a − b = z0 . Note that (x0 , y0 ) = 1
04
04
02
and so we have z0 = x0 − y0 , which is a contradiction since it would be a solution to our initial
equality where x00 < x0 .
y02 = 2(2ab) ∗ (a2 − b2 ) = 4ab(a2 − b2 ) ⇔ (
CASE 2: Let z0 be even and y02 be odd. Then, we have the following equalities, where (p, q) = 1
and one is even and the other is odd:
z0 = 2pq
= p2 − q 2
= p2 + q 2 .
y02
x20
Therefore, (x0 y0 )2 = (p2 − q 2 )(p2 + q 2 ) = p4 − q 4 but since x0 y0 is odd, then we are back to CASE
1.
Problem 2
By contradiction, assume that there exists a solution to the equality x4 − y 4 = 2z 2 . Denote this solution (x0 , y0 , z0 ) and w can choose it so that all terms are pairwise coprime and x0 is the smallest
possible x.
Observe that the parity of x0 and y0 doesn’t change when we look at x40 and y04 instead. Since the
right hand-side is even, we know that both x0 and y0 have the same parity. However, note that if
they are both even, then we could look at the solution (x0 /2, y0 /2, z0 /4). But this is a contradiction
since this solution would have a smaller value for x. Hence, x0 and y0 must both be odd.
Now, as suggested by the hint, notice that x4 − y 4 = (x2 − y 2 )(x2 + y 2 ) = (x − y)(x + y)(x2 + y 2 ).
Hence we can look at the following equation: (x0 − y0 )(x0 + y0 )(x20 + y02 ) = 2z02 . And since x0
and y0 are both odd, then all terms on the left-hand side are even. Hence, we can rewrite:
(x0 − y0 ) = 2a,
(x0 + y0 ) = 2b,
(x20 + y02 ) = 2c
Therefore, we obtain:
8abc = 2z02 ⇒ abc = (
z0 2
)
2
But notice that a, b, c are pairwise coprime. Therefore each one of them has to be a perfect square.
Thus, we have
(x0 − y0 ) = 2a02 ,
(x0 + y0 ) = 2b02 ,
(x20 + y02 ) = 2c02 .
Now, using substitution it follows that:
(x20 + y02 ) = 2c02
2x20 + 2y02 = 4c02
(x0 − y0 )2 + (x0 + y0 )2 = 4c02
4a04 + 4b02 = 4c02
a04 + b02 = c02
We know that this last equality has no solution. Therefore, the initial equation x4 − y 4 = 2z 2 has
no solution.
Problem 3
We can look at this |2x − 3y | = 1 as two different equalities: (1) 2x − 3y = 1 and (2) 2x − 3y = −1.
CASE (1): Consider 2x − 3y = 1. We can try a few small values of x and see if we can find some
solutions:
(1) Let x = 0, then we have 1 − 3y = 1 ⇒ 3y = 0. Impossible!
(2) Let x = 1, then we have 2 − 3y = 1 ⇒ 3y = 1 ⇒ y = 0. One solution (1, 0)
(2) Let x = 2, then we have 4 − 3y = 1 ⇒ 3y = 3 ⇒ y = 1. One solution (2, 1)
Now, we can show that there is no solution for x ≥ 3. Let x ≥ 3:
Then note that 2x ≡ 0 (mod 8). Hence we have 1 ≡ 2x − 3y ≡ −3y (mod 8). Let’s look at the
congruences of −3y modulo 8.
−30 ≡ −1 (mod 8)
−31 ≡ −3 (mod 8)
−32 ≡ −1 (mod 8)
−33 ≡ −3 (mod 8)
Therefore, we can see that −3y ≡ −1, −3 (mod 8), but we will never have −3y ≡ 1 (mod 8).
Hence, there is no solution for x ≥ 3. The two solutions to 2x − 3y = −1 are (1, 0) and (2, 1).
CASE (2): Consider 2x − 3y = −1. We can try a few small values of x and see if we can find some
solutions:
(1) Let x = 0, then we have 1 − 3y = −1 ⇒ 3y = 2. Impossible!
(2) Let x = 1, then we have 2 − 3y = −1 ⇒ 3y = 3 ⇒ y = 1. One solution (1, 1)
(2) Let x = 2, then we have 4 − 3y = −1 ⇒ 3y = 5.Impossible
Now, we can show that there is no solution for x ≥ 3. Let x ≥ 3:
Then note that 2x ≡ 0 (mod 8). Hence we have −1 ≡ 2x − 3y ≡ −3y (mod 8). Based on the
congruences computed in Case (1), we know that 3y ≡ 1 (mod 8) when y = 2k. Then we get:
32k−1 = (3k − 1)(3k + 1) = 2r where the two terms differ by two and r ∈ N. Thus, we can
write (3k − 1) = 2s and (3k + 1) = 2t where s + t = r. The only values of s and t for which
2s and 2t differ by 2 are (s, t) = (1, 2). Hence, the only solution would be 3k − 1 = 21 = 2 and
3k + 1 = 22 = 4, that is, k = 1 and so y = 2k = 2 implying that x = 3.
Therefore, for 2x − 3y = −1 there are exactly two solutions: (1, 1) and (3, 2).
Conclusion: There are 4 solutions for the equation |2x − 3y | = 1. They are (1, 0), (2, 1), (1, 1), and
(3, 2).
Problem 4
Consider y 2 = x3 + 7. As suggested in the hint, add 1 on both sides to get y 2 + 1 = x3 + 8. Now,
consider both sides of this equality modulo 4:
y = 0 ⇒ y 2 − 1 ≡ 3 (mod 4)
x = 0 ⇒ x3 ≡ 0 (mod 4)
y = 1 ⇒ y 2 − 1 ≡ 0 (mod 4)
x = 1 ⇒ x3 ≡ 1 (mod 4)
y = 2 ⇒ y 2 − 1 ≡ 3 (mod 4)
x = 2 ⇒ x3 ≡ 0 (mod 4)
y = 3 ⇒ y 2 − 1 ≡ 0 (mod 4)
x = 3 ⇒ x3 ≡ 3 (mod 4)
We have two common values 0 and 3. But we can start by observing that if x is even, then
y 2 ≡ x3 + 7 (mod 8) ≡ 7 (mod 8). This is impossible since 7 (mod 8) is not a square. So x has to
be odd.
We need y is even and x ≡ 3 (mod 4); then both sides are congruent to 3 modulo 4. Note that we
can rewrite x3 + 8 = x3 + 23 = (x + 2)(x2 − 2x + 4); hence, y 2 − 1 = (x + 2)(x2 − 2x + 4). But
since x ≡ 3 (mod 4), then x2 − 2x + 4 = (x − 1)2 + 3 ≡ 22 + 3 ≡ 3 (mod 4). Thus, x2 − 2x + 4
has to be divisible by a prime factor p ≡ 3 (mod 4). This implies that p also divide y 2 + 1; then
y 2 ≡ −1 (mod p) and so p ≡ 1 (mod 4). This is a contradiction to the fact that p ≡ 3 (mod 4).
Therefore, there is not solution to this equation.
Problem 5
(⇒) Assume that x ∈ N is expressible as the difference of 2 squares. Then we have x =
a2 − b2 = (a − b)(a + b). Since (a − b) and (a + b) must have the same parity, we know that
(a − b)(a + b) 6≡ 2 (mod 4).
(⇐) Assume that x ∈ N such that x 6≡ 2 (mod 4).
Case 1: let x be odd, then we have x = (2k − 1) = (k + 1)2 − k 2 .
Case 2: let x ≡ 0 (mod 4), then we have x = 4k = (k+1)2 −(k−1)2 = k 2 +2k+1−k 2 +2k−1 =
4k.
Problem 6
As suggested by the hint, we can write n − 169 = a2 + b2 + c2 + d2 since any natural number can
be represented as the sum of 4 integer squares (but some can be zero). Now, we want to show that
we can always find a, b, c, and d non-zero.
(1) Assume that none of a, b, c, d is 0. Then we have
n − 169 = a2 + b2 + c2 + d2 ⇒ n = a2 + b2 + c2 + d2 + 132 .
(2) Assume that exactly one of a, b, c, d is 0, say d. Then we have
n − 169 = a2 + b2 + c2 ⇒ n = a2 + b2 + c2 + 132 = a2 + b2 + c2 + 52 + 122 .
(3) Assume that exactly two of a, b, c, d are 0, say c and d. Then we have
n − 169 = a2 + b2 ⇒ n = a2 + b2 + 133 = a2 + b2 + 32 + 42 + 122 .
(4) Assume that exactly three of a, b, c, d are 0, say b, c and d. Then we have
n − 169 = a2 ⇒ n = a2 + 132 = a2 + 42 + 42 + 42 + 112 .
Therefore, any integer n > 169 is expressible as the sum of 5 squares of positive integers.