$a, b \in\Bbb N$, $A=\sqrt{a^2+2b+1}+\sqrt{b^2+2a+1}\in \Bbb N $, to show that $a = b$. If $a$ and $b$ are positive integers such that $A=\sqrt{a^2+2b+1}+\sqrt{b^2+2a+1}\in \Bbb N $, to show that $a = b$.
By contradiction assuming that $a <b$ then it follows that $2 (a +1) <A <2 (b +1)$.
This means that $A = 2a + r,$ $r$ taking values  $3, 4, ..., 2b-2a +1$.
From the equality 
(1) $\sqrt{a^2+2b+1}+\sqrt{b^2+2a+1}= 2a+r$ following
$(a-b)(b+a-2)=(2a+r)(\sqrt{a^2+2b+1}-\sqrt{b^2+2a+1})$ and then
(2) $\sqrt{a^2+2b+1}-\sqrt{b^2+2a+1}=\frac{(a-b)(b+a-2)}{2a+r} \in \Bbb Q. $
From (1) and (2) obtain $\sqrt{a^2+2b+1}\in \Bbb Q$ and $ \sqrt{b^2+2a+1}\in \Bbb Q$. Finally
$a^2+2b+1=p^2$ and $b^2+2a+1= q^2$, $p, q \in \Bbb N.$
Hence the attempts made ​​not lead to anything conclusive. Does anyone have any suggestions?
 A: First, if $n\in \Bbb N$ and $\sqrt{n} \in \Bbb Q$, then $\sqrt{n} \in \Bbb N$.
To see why, notice first that if $n$ is a perfect square, there is nothing to prove.
So, suppose $n\in \Bbb N$ and $\sqrt{n} \in \Bbb Q$, and $n$ is not a perfect square. Then there is a prime number $\alpha$ such that $\alpha$ factors in $n$ to an odd power, that is: $\alpha^{2k-1}|n$ but $\alpha^{2k}\not|n$, for some integer $k$.
You have $\sqrt{n}=\frac{p}{q}$, with coprime $p$ and $q$. Then $p^2=nq^2$. Since $\alpha^{2k-1}|n$, you have also $\alpha^{2k-1}|p^2$, thus $\alpha^{k}|p$, but then $\alpha^{2k}|nq^2$, thus $\alpha^{2k}|n$ or $\alpha|q$, but both are impossible.
Hence there is a contradiction, and $n$ must be a perfect square, so $\sqrt{n} \in \Bbb N$.
Edit: There was a mistake above, $\alpha^2$ may divide $n$, the important thing it that there is a prime $\alpha$ with an odd power in the prime factorization of $n$, when $n$ is not a perfect square. For example, $27$ is not a square, but obviously $3^2|27$.

Now, there is no square integer between $b^2$ and $b^2+2b+1=(b+1)^2$. Since $a<b$, $b^2+2a+1$ can't be a perfect square, so its square root can't be a rational, thus you have the contradiction you want.
