1. Proofs-squares

prove that the sum of two odd squares cannot be a square

2. Originally Posted by anncar
prove that the sum of two odd squares cannot be a square
Let the two odd numbers be $\displaystyle x = 2m + 1$ and $\displaystyle y = 2n + 1$

Sum the two squares and note that the result is divisible only by 2, not 4.

Now let any even number be $\displaystyle z = 2p$. The square of this number is divisible by 4.

etc, etc.

-Dan

3. Hello, anncar!

A variation of Dan's explanation . . .

Prove that the sum of two odd squares cannot be a square.

If $\displaystyle n$ is even, $\displaystyle n = 2k$, its square is: .$\displaystyle n^2 \:=\:(2k)^2\:=\:4k^2$
. . The square of an even number is a multiple of 4. .[1]

If $\displaystyle n$ is odd, $\displaystyle n = 2k+1$, its square is: .$\displaystyle n^2\:=\:(2k+1)^2 \:=\:4(k^2 + k) + 1$
. . The square of an odd number is one more than a multiple of 4. .[2]

These are the only two forms for the square of an integer.

Let the odd integers be: .$\displaystyle a \:=\:2h+1,\;b\:=\:2k+1$

The sum of their squares is: .$\displaystyle N \;=\;a^2+b^2 \;=\;(2h+1)^2+(2k+1)^2$

. . $\displaystyle =\:4h^2+4h + 1 + 4k^2 + 4k + 1 \:=\:4(h^2+h+k^2+k)+2$

Hence, $\displaystyle N$ is two more than a multiple of 4.

Since $\displaystyle N$ cannot be either of the forms [1] or [2], $\displaystyle N$ cannot be a square.