<s>[INST] $a-b \nmid a^2+b^2$ if $a>b+2$ and $\gcd(a,b)=1$ 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?
 [/INST] 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.
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