4

Any two co-prime number $a,b$ with $a>b+2$ we have $a^2+b^2$ is not divisible by $a-b$, $a,b \in \mathbb{N}$. But how to prove this?

TRUSKI
  • 1,113
  • Negating $b$ in the dupe proofs yields what you desire. – Bill Dubuque Jun 26 '17 at 12:44
  • Here the result follows form your question but this is not exactly what your question was.! – TRUSKI Jun 26 '17 at 12:48
  • The dupe (with $b$ negated) shows $,(a-b,a^2+b^2)\mid 2,$ so if $,a-b\mid a^2+b^2$ then the gcd $= a-b \mid 2,,$ contra $a-b > 2.,$ That $,1= (a,b),\Rightarrow, 1= (a,b^2)=(a,a^2+b^2),$ is immediate from Euclid. By symmetry $(b,a^2+b^2)=1.,$ This and related results are also proved in many other questions. Please search before posing common questions. – Bill Dubuque Jun 26 '17 at 12:56
  • I got that. I searched but I did not get. So I had the post. Thanks for your comments. – TRUSKI Jun 26 '17 at 13:01
  • You can search using approach0, e.g. this search. – Bill Dubuque Jun 26 '17 at 13:02

2 Answers2

1

If $a-b \mid a^2+b^2$ then $\gcd(a^2+b^2,a-b)=a-b \ge 3$. Now $$ \gcd(a^2+b^2,a-b)=\gcd(a^2+b^2-(a-b)^2,a-b)=\gcd(2ab,a-b). $$ Since $\gcd(a,b)=1$ then every prime dividing $a$ or $b$ cannot divide $a-b$. Hence $$ \gcd(a^2+b^2,a-b) \in \{1,2\}. $$ In particular, your conjecture is true.

Paolo Leonetti
  • 15,423
  • 3
  • 24
  • 57
0

Since $a-b$ divides $a^2-b^2$, if it also divides $a^2+b^2$ then $a-b$ divides $2b^2$.

Now use that $a$ and $b$ are coprime to see that $a-b$ and $b^2$ are coprime. So by Euclides' lemma $a-b$ divides $2$. This contradicts the condition $a-b>2$.

ajotatxe
  • 65,084