Prove that if $ b> a $ Let $ a $ be a positive integer and $ b $ a divisor d $ a ^ 2 + 1 $. Prove that if $ b> a $. So $ b> a + \sqrt{a}$
What did I do:
Let b = a + k for positive integer k.
We have to prove that $ k > \sqrt{a}$
$(a^2+1)/(a+k)=a-k+(k^2+1)/(a+k)$. Thus,
$(k^2+1)/(a+k)$ has to be whole number and it is obvious it is positive.
That means $ k^2+1 > a+k$
$k^2-k+1 > a$
But if $k <= \sqrt{a} $ then $k^2 <= a$ and $k^2 - k + 1 <= a$, contradiction.
Are there other methods to prove it?
 A: The statement you're asking to prove is not always true. For example, if $a = 1$, then $a^2 + 1 = 2$, so $b = 2$ is a divisor. However, you then have $a + \sqrt{a} = 1 + 1 = 2 = b$. However, changing the $\gt$ to $\ge$ will work, i.e., $b \ge a + \sqrt{a}$.
Your proof is basically correct, but you made a small mistake. You stated that $\frac{k^2 + 1}{a + k}$ is a positive whole number. As such, it could be $1$, so the proper inequality to use is
$$k^2 + 1 \ge a + k \implies k^2 - k + 1 \ge a \tag{1}\label{eq1}$$
Thus, if $k \lt \sqrt{a}$, you have $k^2 \lt a \implies k^2 - k + 1 \lt a$, which is now a contradiction.
Note the only edge case occurs when $k = 1$, giving $a = 1$ and $b = 2$, which I stated originally. Thus, to keep the original condition of $b \gt a + \sqrt{a}$, you can just simply add the extra condition that $a \gt 1$.
As for other methods to prove it, there are a few, but the technique you used is the one I would've also used, and I believe it's at least one of the simplest & most direct ways to prove this.
