# How to prove that there's at least one of them to be multiple 5?

Mathematics Asked by Henry Cai on December 2, 2020

how do we prove that if $$a^2+b^2=c^2$$, and they are all integers, there would be at least one of them to a multiple of 5.
What I did is to assume that none of them to be multiple of 5. So$$a=5p_1+r_1, b=5p_2+r_2, c=5p_3+r_3, 0.
By direct substitution back to the equation, we would have:
$$5(5p_1^2+2p_1r_1+5p_2^2+2p_2r_2-5p_3^2-2P_3r_3)=r_3^2-r_1^2-r_2^2$$.
Then, we can list all possible scennarios to show that for this equality to be held, there must be at least 1 r to be 0. I want to ask is there any quick way we can show $$r_3^2-r_1^2-r_2^2$$is a multipler of 5 unless one of them to be 0. Thank you.

Modulo $$5$$, all squares are in $${-1,0,1}$$. The only ways two squares can add up to a third square modulo $$5$$ are $$0+0=0$$ $$1+0=1$$ $$-1+0=-1$$ $$1+(-1)=0$$ There is a zero in all four possibilities. So in $$a^2+b^2=c^2$$, at least one variable is a multiple of $$5$$.