<s>[INST] If $a>1$ and $b \mid a^2+1$ and $b>a$.Prove that $b-a > \sqrt{a}$. If $a>1$ and $b \mid a^2+1$ and $b>a$.Prove that $b-a > \sqrt{a}$.
I take $b=a+m$ where $m$ is a natural number then we have:
$b \mid a^2+1 \Rightarrow b \mid m^2+1 \Rightarrow b-1 \le m^2 \Rightarrow a \le m^2 \Rightarrow \sqrt{a} \le b-a$
But I don't know why I get the equality case with it.I can't even prove or disprove that there is an equality case.
 [/INST] With $b=a+m$ we have $b| b(a-m)=(a+m)(a-m)=a^2-m^2=(a^2+1)-(m^2+1),$ so $b|m^2+1,$ so $b\leq m^2+1.$
Now if $m\leq \sqrt a$ then $a<b\leq m^2+1\leq (\sqrt a)^2+1=a+1 ,$ implying $b=a+1.$ But $a+1$ cannot divide $a^2+1.$ Otherwise $a+1| a(a+1)-(a^2+1)=a-1>0,$ which implies $a+1\leq a-1,$ a paradox. Therefore $m>\sqrt a.$
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