Let $x+y=mn,x-y=m^2-n^2$ , where $m,n\in N^*$ $x^2+y^2-[8(x+y)+2(xy+1)]$ $=(x-y)^2-8(x+y)-2$ $=m^4+n^4-2m^2n^2-8mn-2$

I have two doubts regarding your solution: 1. The equation in integers $ab=c^2$ does not imply that $a$ and $b$ are both squares, unless $\text{gcd}(a,b)$ is a square itself. So why do you assume $m,n \in \mathbb{N}_0$? 2. You reduced the original inequality to $(m^2-n^2)^2>8mn+2$ for some distinct real $m,n$.. Probably I am missing something, but how does the conclusion follow? Cheers, Paolo L.

Here is my solution: $z^2+z(x+y)+xy-x^2-y^2-2xy=0 \implies z^2+z(x+y)-x^2-y^2-xy=0$ $\implies 5x^2+5y^2+6xy=t^2$, with $t \in \mathbb N$ $\implies (2(x+y))^2+(x-y)^2=t^2$ $\implies t=2(x+y)+r,r>0$ $\implies 5x^2+5y^2+6xy=4x^2+4y^2+8xy+4r(x+y)+r^2$ $\implies x^2+y^2=2xy+4r(x+y)+r^2$ Now we only need to show that $4r(x+y)+r^2>8(x+y)+2$ i.e. $r>1$ If $r=1$, then $x^2+y^2=2xy+4(x+y)+1$ $\implies x^2-x(2y+4)+y^2-4y-1=0$ $\implies 32y+20=k^2,k \in \mathbb N$, impossible since $\left(\frac{5}{8}\right)=-1$. So the inequality is proved.

sqing wrote: Let $x,y,z$ be distinct positive integers such that $(y+z)(z+x)=(x+y)^2$ . Show that \[x^2+y^2>8(x+y)+2(xy+1).\] (Paolo Leonetti) I will prove more sharper inequality , that is $x^2+y^2 \ge 8(x+y)+2(xy+2)$ . This is equivalent to $(x-y)^2 \ge 8(x+y)+4$ . So now making a quadratic from 1st equation , and since it has integer roots , we must have discriminant($D$) as perfect square . We get $D = 5x^2+5y^2+6xy$ . Now, take $t=x-y \implies y=x-t$ putting in equation gives $D = 16x^2-16x+5t^2 = (4x-2t)^2+t^2$ . Obviously , we must have $D \ge (4x-2t+1)^2$ . But if $D = (4x-2t+1)^2$ , then we get $8x = (t+2)^2-5$ a contradiction . Thus , $D \ge (4x-2t+2)^2$ which gives $t^2 \ge 16x-8t+4 \implies (x-y)^2 \ge 8(x+y)+4$ . So we are done $\Box$ ^^ please point out the errors if any.

That looks correct, but I also proved it in my previous post: In $ x^2+y^2=2xy+4r(x+y)+r^2 $, I proved that $r \geq 2$, hence $x^2+y^2 \geq 8(x+y)+2(xy+2)$, as needed.