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Prove if $a$ and $b$ are odd then $a^2+b^2$ is not a perfect square.

We have been learning proof by contradiction and were told to use the Euclidean Algorithm.

I have tried it both as written and by contradiction and can't seem to get anywhere.

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All squares are congruent to 1 or 0 mod 4. If they are odd they are congruent to 1 mod 4. therefore the sum of two odd squares is congruent to 2 mod 4. Thus not a square.

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  • $\begingroup$ How do we prove perfect squares are 0 mod 4 or 1 mod 4 then? $\endgroup$
    – user129818
    Feb 20 '14 at 17:27
  • $\begingroup$ let $a=2n\rightarrow a^2=4n^2\equiv 0 \bmod 4$ $\endgroup$
    – Yorch
    Feb 21 '14 at 2:43
  • $\begingroup$ let $a=(2n+1)^2=4n^2+4n+1=4(n^2+n)+1\equiv1 \bmod 4$ $\endgroup$
    – Yorch
    Feb 21 '14 at 2:44
  • $\begingroup$ @user129818 Or check all of the four cases $a\equiv 0 \pmod 4$, $a\equiv\pm 1\pmod 4$, and $a\equiv 2\pmod 4$ manually. We have $0^2=0$, and $(\pm 1)^2 = 1$, and $2^2=4\equiv 0$, so all of them are either zero or one modulo four. $\endgroup$ Dec 24 '18 at 12:31
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Here is a different way of doing this.

Suppose $a^2+b^2=c^2$ with $a$ and $b$ odd, then $c^2$ is even, so must be divisible by $4$.

Now consider the even square $(a+b)^2=c^2+2ab$. The first two terms are divisible by $4$ while $2ab$ is twice an odd number, so the equation can't hold.


One can also go to $(a+b+c)(a+b-c)=2ab$ with both factors on the left being even, but the right-hand side not divisible by $4$.


Note also that $(2m+1)^2=8\left(\dfrac {m(m+1)}{2}\right)+1\equiv 1 \bmod 8$, and though working modulo $4$ is enough here, this stronger fact is useful to know.

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Let $a=2m+1$ and $b=2n+1$.

Assume $a^2+b^2=k^2$. Then: $$(2m+1)^2+(2n+1)^2=k^2 \iff \\ 4(m^2+m+n^2+n)+2=k^2 \iff \\ 4(m^2+m+n^2+n)+2=(2r)^2 \iff \\ 2(m^2+m+n^2+n)+1=2r^2 \iff \\ 2s+1=2r^2,$$ which is a contradiction. Hence, the assumption $a^2+b^2=k^2$ is false.

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