Pythagorean Triples in Disguise?

We must have $y≥0$ .As the right hand side is non-zero,so must be the left hand side,hence $|x|≥|y|+1$ or $|x|≤|y|-1$ .

In either case $(x^{2}-y^{2})^{2}≥(2y-1)^{2}$ ,so $(2y-1)^{2}≤1+16y$ .

Solving the inequation we get $y≤5$ .Trying all such values of $y$ yield the solutions.

I will be writing a little long solution .

From the given question $16y+1$ is a perfect square. So we take $16y+1={ n }^{ 2 }$ where $n\epsilon I$ . Our equation becomes ${ x }^{ 2 }-{ y }^{ 2 }=\pm n$

$\Rightarrow 16y=(n+1)(n-1)$ .

Now $n$ can't be even.

Because then $(n+1),(n-1)$ would become odd.

Also $n$ is not of the form $8k+3(k\epsilon I)$ because then

$16y=(8k+2)(8k+4)\quad \Rightarrow 2y=(4k+1)(2k+1)\quad$ Hence not possible.

So $n$ is either of the form $8k+1,8k-1$ .

Let the counting begin.

The values of $n=1,7,9,15,17,23,25.........$

The corresponding values of $y=0,3,5,14,18,33,39....$

Solving for $x$ $\\ y=0,x=1\quad \\ y=3,x=4\quad ({ x }^{ 2 }-9=\pm 7)\\ y=5,x=4\quad (we\quad are\quad taking\quad either\quad the\quad positive\quad or\quad negative\quad n\quad in\quad the\quad \\ \quad \quad \quad \quad \quad \quad \quad \quad equation\quad { x }^{ 2 }-{ y }^{ 2 }=\pm n\quad for\quad getting\quad solutions\quad of\quad x)$

We have to now prove that for $y\ge 14$ no positive integer x exists.

$Case-1\quad x\ge y+1\quad (For\quad the\quad equation\quad { x }^{ 2 }-{ y }^{ 2 }=n)\\ Now\quad { x }^{ 2 }\ge { y }^{ 2 }+2y+1\\ \Rightarrow { y }^{ 2 }+n\ge { y }^{ 2 }+2y+1\\ \Rightarrow n\ge 2y+1\\ \Rightarrow { n }-1\ge ({ n }^{ 2 }-1)/8\quad \quad So\quad 0\ge { n }^{ 2 }-8n+7\quad hence\quad n\epsilon [1,7]\\ \Rightarrow y\quad can\quad be\quad 0,3\\ Case-2\quad x\le y-1\quad (For\quad the\quad equation\quad { x }^{ 2 }-{ y }^{ 2 }=-n\quad )\\ Now\quad { x }^{ 2 }\le { y }^{ 2 }-2y+1\\ \Rightarrow { y }^{ 2 }-n\le { y }^{ 2 }-2y+1\\ \Rightarrow ({ n }^{ 2 }-1)/8\le n+1\quad \quad So\quad { n }^{ 2 }-8n-9\le 0\quad hence\quad n\epsilon [-1,9]\\ \Rightarrow y\quad can\quad be\quad 0,3,5\\ So\quad we\quad have\quad proved\quad that\quad these\quad are\quad the\quad only\quad solutions.\quad \\ \\$

Hence $\sum _{ a=1 }^{ n }{ { x }_{ a }+{ y }_{ a } } =17$ Please notify me about any typing mistakes since it was a long solution

Solve Diophantus and sum is 17.