Question

Let x, y, z be a primitive Pythagorean triple with y even. Prove that x+y ≡ x−y ≡ ±1 mod 8.

Answer 1

Since is even, we know that one of the following must hold:

Note that this means

Thus, either , or .

Since is primitive, they have no common divisor other than . Thus, at least one of must be odd (else will divide all three). Say is odd; then is odd too, and hence,

which implies must be odd. Similarly, if is odd, then so is , and hence,

which implies must be odd.

Thus, we get that are odd and is even.

Since is odd, we know that one of the following must be true:

This means

Thus, in any case, we have , just from the fact that is odd. Similarly, because is odd. On the other hand, we have seen at the beginning that or . If then we get

which is impossible since and . Therefore, which means .

Suppose that . Then and . On the other hand, when , we get and . Thus, in any case, we have

Now, we have

which means . But then, we get

and similarly,

Thus, we have got and .

Now recall that, since is primitive Pythagorean triple, there are integers such that exactly one of them is even, and

Thus, . Note that is odd, and hence, (as seen before in case of odd). If is even then (as seen before in case of even), and hence, , which implies

If is odd then (as seen before in case of odd), and hence, , which implies

.

Thus, we have shown the following:

a)

b) either

or

These are all we were required to prove.