please clr my doubt $y^2-y+x^2$ - it is a quadratic over $y$. Now for perfect square condition discriminant $D$ =0 .It implies that $1-4x^2=0$ giving x^2=1/4 ?

@above The problem states that $x^2 + y^2 -y$ is a square of a "natural number", not a square of a "polynomial of $y$".

$(x,y)=(2,4)$ give us $x^2+y^2-y$ is square and $4+2+1=7$ is prime. $(x,y)=(13,17)$ give us $31$ prime. $(x,y)=(17,96)$ give us $113$ prime. Indeed there are many other counterexamples. Please check problems correctly

@above I checked the documents and I realized that there was a typo in the expression. I edited it.

SinaQane wrote: The natural numbers $x, y > 1$, are such that $x^2 + xy -y$ is the square of a natural number. Prove that $x + y + 1$ is a composite number. Suppose for the sake of contradiction $x+y+1$ is a prime number. Let $x^2+xy-y=n^2$ and given $x,y>1$ we have $n>1$. So notice $$x^2+xy-y=x(x+y)-y\equiv -(x+y)\equiv 1\equiv n^2\pmod{x+y+1}$$So $x+y+1\mid (n-1)(n+1)$. i.e $x+y+1\mid n\pm 1$. If $x+y+1\mid n+1$ we get $n\ge x+y$. But so $x^2+xy-y=n^2\ge x^2+2xy+y^2$ so we have $-y\ge xy+y^2$ contradiction. Similarly if $x+y+1\mid n-1$ we get $n>x+y+1$ and so $x^2+xy-y=n^2>x^2+2xy+y^2+2x+2y+1$ clearly contradiction. Thus $x+y+1$ can't be a prime.

Let $x^2+xy-y=k^2$, then $y(x-1) = k^2-x^2$, so $y=\frac{k^2-x^2}{x-1}$. Since $x-1\mid k^2-x^2$, we may write $x-1=b_1b_2$ where $b_1\mid k-x$ and $b_2\mid k+x$. Now, \[x+y+1 = x+ \frac{k^2-x^2}{x-1}+1 = \frac{k^2-1}{x-1}= \frac{k-1}{b_1} \cdot \frac{k+1}{b_2} = \left(\frac{k-x}{b_1} +\frac{x-1}{b_1}\right)\left(\frac{k+x}{b_2} +\frac{-x+1}{b_2}\right)\]Thus, both terms $\frac{k-1}{b_1} ,\frac{k+1}{b_2}\in \mathbb{Z}$. Now, we are left to verify that neither can equal 1. Note that both $b_1,b_2\leq b_1b_2 = x-1 < k-1$ since $y$ is positive, so both terms are $>1$ and we're done. $\blacksquare$.

Obviously we have $x + y + 1 > 2$. Assume that $x + y + 1$ is prime and $z^2 = x^2 + xy - y$ for some $z \in \mathbb{N}$. We have $(x+y+1)(x-1) = z^2 - 1 = (z-1)(z+1)$. Since $x+y+1$ is prime and greater than $2$, it follows that it divides $z \pm 1$ which implies $x+y+1 \leq z+1$. But note that $z < x+y$ since $z^2 = x^2 + xy - y < x^2 + 2xy + y^2$, so this gives $x+y+1 \leq z+1 < x+y+1$, a contradiction.