$4n( N-n) +1=s^{2} \ \ \ \Longleftrightarrow \ n^{2}\left( N^{2} +1\right) =\left( n^{2} +(( s-1) /2)^{2}\right)\left( n^{2} +(( s+1) /2)^{2}\right)$ Note that $|s|>1$ and $s$ must be an odd integer. Suppose that $N^{2} +1$ is a prime number. Then $n^{2} \geq \min(n^{2} +(( s+1) /2)^{2},n^{2} +(( s-1) /2)^{2}) \geq n^{2} +1$, which is impossible. in more detailsLet $q=N^{2} +1$. If $q\mid n^{2} +(( s-1) /2)^{2}$, then $n^{2} =\left( n^{2} +(( s-1) /2)^{2}\right) /q\cdotp \left( n^{2} +(( s+1) /2)^{2}\right) \geq n^{2} +(( s+1) /2)^{2}$. If $q\mid n^{2} +(( s+1) /2)^{2}$, then $n^{2} =\left( n^{2} +(( s+1) /2)^{2}\right) /q\cdotp \left( n^{2} +(( s-1) /2)^{2}\right) \geq n^{2} +(( s-1) /2)^{2}$. In any case, we have $n^{2} \geq n^{2} +1$, which is impossible. Suppose that $N^{2} +1$ is a composite number. To prove the existence of solution, we will show that the following prop is true : Proposition There exists $( a,b,c,d) \in \mathbb{Z}^{4}$ such that $N^{2} +1=\left( a^{2} +b^{2}\right)\left( c^{2} +d^{2}\right)$, $ad-bc=1$, and $abcd\neq 0$. proof of this propLet $R=\mathbb{Z}[ i]$. $\left( i=\sqrt{-1}\right)$ It is well-known that $R$ is UFD. Since $N^{2} +1$ is a composite number, $N+i$ is neither irreducible element or unit. So there exists $( a,b,c,d) \in \mathbb{Z}^{4}$ such that $N+i=( a-bi)( c+di) \cdots ( \spadesuit )$ and neither $a-bi$ or $c+di$ is units $\cdots ( \clubsuit )$ Since $( a-bi)( c+di) =( ac+bd) +( ad-bc) i$ and $( \spadesuit )$, we have $1=ad-bc\cdots ( \heartsuit )$. Since $(\heartsuit$) and $( \clubsuit $), we can easily verify that $abcd\neq 0\cdots ( \diamondsuit )$. By taking of complex conjugate of $( \spadesuit )$, we have $N^{2} +1=\left( a^{2} +b^{2}\right)\left( c^{2} +d^{2}\right)$. $\blacksquare $ By the proposition, we can take $( a,b,c,d) \in \mathbb{Z}^{4}$ such that $N^{2} +1=\left( a^{2} +b^{2}\right)\left( c^{2} +d^{2}\right)$, $ad-bc=1$, and $abcd\neq 0$. Let $n=|ac|$ and $s=2bc+1$. $n^{2}\left( N^{2} +1\right) =( ac)^{2}\left( a^{2} +b^{2}\right)\left( c^{2} +d^{2}\right)\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\left(( ac)^{2} +( bc)^{2}\right)\left(( ac)^{2} +( ad)^{2}\right)\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\left( n^{2} +(( s-1) /2)^{2}\right)\left( n^{2} +(( s+1) /2)^{2}\right) $ Note that we have $n< N$ because $N^{2} -n^{2} =a^{2} d^{2} +b^{2} c^{2} +b^{2} d^{2} -1 >0$. Therefore, there exists $n\in \mathbb{Z}^{+}$ such that $n< N$ and $4n( N-n) +1$ is a perfect square. $\blacksquare $