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$$a=5p_1+r_1, b=5p_2+r_2, c=5p_3+r_3, 0<r<5$$.
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.
Asked
Active
Viewed 42 times
1
Henry Cai
- 633
-
Hint : Try all the cases according to the class of $a$, $b$, $c$ mod $5$. – TheSilverDoe Nov 08 '20 at 12:55
-
I see. thank you. I am still quite confusing about modular arithmetic.. Haven't learnt it before – Henry Cai Nov 08 '20 at 13:02
-
Please read Pythagorean Triples. One is always divisible by $5$ and one is always divisible by $3$. – Math Lover Nov 08 '20 at 13:20
-
One more you can go through - https://math.stackexchange.com/questions/1154424/observations-on-integer-sided-right-triangles – Math Lover Nov 08 '20 at 13:21
1 Answers
1
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 $5$.
Parcly Taxel
- 103,344
-
Thank you very much for your answer. But, how do we show that all squares are in {-1,0,1}, thank you. – Henry Cai Nov 08 '20 at 13:02
-
-
-
are we saying about the squares of r or the squares of every integers. If for every integers...we can't actually brute force....? – Henry Cai Nov 08 '20 at 13:07
-
@HenryCai Because we are taking modulo $5$, we only need to test for $5$ consecutive integers; the other integers will just repeat. Conventionally we'd take $0,1,2,3,4$ as the test integers. – Parcly Taxel Nov 08 '20 at 13:22