$\sqrt{x+1}+\sqrt{y+1}$ and $\sqrt{x-1}+\sqrt{y-1}$ are non-consecutive integers We have
$$a=\sqrt{x+1}+\sqrt{y+1}$$  $$b=\sqrt{x-1}+\sqrt{y-1}$$ $$x,y>0$$
And we have to find $x$ and $y$ such that $a$ and $b$ are non-consecutive integers. One solution may be 5/4 for both, $x$ and $y$, but I have no idea how to show this.
 A: Assume that $a=b+n$ for an integer $n\geqslant2$, then
$$
\frac1{u(x)}+\frac1{u(y)}=\frac{n}2\qquad \mbox{with}\qquad
u(z)=\sqrt{z+1}+\sqrt{z-1}.
$$
For every $z\geqslant1$, $u(z)\geqslant\sqrt2$ hence the LHS is $\leqslant\sqrt2$. Thus $n\geqslant2$ and $n\leqslant2\sqrt2<3$. Since $n$ is an integer, $n=2$. Since $x\geqslant1$ and $y\geqslant1$, one can represent $(x,y)$ as $x=v(s)$ and $y=v(t)$ for some $s$ and $t$ in $]0,1/\sqrt2]$, where the function $v$ is defined by
$$
v(z)=z^2+1/(4z^2).
$$
A simple computation yields $a=s+1/(2s)+t+1/(2t)$ and $b=1/(2s)-s+1/(2t)-t$ hence $a=b+2(s+t)$ and one must choose $s$ and $t$ such that $s+t=1$. The general solution to this is 
$$
(x,y)=(v(s),v(1-s)),\qquad 1-1/\sqrt2\leqslant s\leqslant1/\sqrt2,
$$
which yields
$$
a=1+1/(2s(1-s)),\quad b=-1+1/(2s(1-s)).
$$
The condition that $a$ and $b$ are integers is met for every $s$ such that $2s(1-s)=1/k$ with $k\geqslant1$ an integer. The condition that $1-1/\sqrt2\leqslant s\leqslant1/\sqrt2$ implies that $\sqrt2-1\leqslant2s(1-s)\leqslant1/2$. 
Since $1/3\lt\sqrt2-1$, the unique solution is $k=2$, that is, $s=1/2$.
Finally, $v(1/2)=5/4$ hence the unique solution is $(x,y)=(5/4,5/4)$, which yields $(a,b)=(3,1)$.
