can anybody please guide me on this
Prove: the sum of the squares of two odd integers cannot be a perfect square, i.e. if x and y are both odd, then there is no perfect square z^2 such that
x^2 + y^1 = z^2.
Thanks
can anybody please guide me on this
Prove: the sum of the squares of two odd integers cannot be a perfect square, i.e. if x and y are both odd, then there is no perfect square z^2 such that
x^2 + y^1 = z^2.
Thanks
Call these two odd integers and , where m and n are both integers. Then the sum of their squares is
This will only be a perfect square if the discriminant is 0, so
But m and n both have to be integers, which means it is impossible for .
Therefore there is no way for the sum of squares of two odd integers to also be a perfect square.
You could also note that if x and y are odd and x^2 + y^2 = z^2, then z is even and therefore z^2 is divisible by 4. However, x^2 + y^2 gives the remainder 2 when divided by 4, as the second line in the first formula of post #2 shows.
to amplify emakarov's post:
note that an odd number is either of the form: 4n + 1, or 4n + 3, that is equal to 1 or 3 (mod 4).
note as well that 1^{2} = 3^{2} = 1 (mod 4).
therefore, the sum of the squares of two odd integers is:
1 + 1 = 2 ≠ 0 (mod 4)
The simplest solution is to use mods, just like emakarov said. Any odd square is going to be 1 mod 4, and 1+1 = 2, so we need a perfect square that is 2 mod 4. However this is impossible since all even squares are 0 mod 4 (this is very easy to show) so we're done.