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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.

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You can check that $x=y=5/4$ actually is a solution by simply doing the arithmetic. Are you asking how one might discover such a solution in the first place? –  Brian M. Scott Nov 15 '11 at 20:54
    
Yes, I was asking how to discover them. I found this solution by simply trying many possible ones (that I could think of). –  Robert Badea Nov 16 '11 at 8:17
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1 Answer 1

up vote 5 down vote accepted

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 $(a,b)=(3,1)$.

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But $a$ and $b$ are supposed to be integers. –  Gerry Myerson Nov 15 '11 at 22:11
    
It doesn't look like $v$ is onto $[1,+\infty)$ on $(0,1/2]$ - it looks like the minimum value is $5/4$. Is there some additional reason $\xi$ and $\eta$ exist? –  Thomas Andrews Nov 15 '11 at 22:15
    
@Gerry, see revised version. –  Did Nov 15 '11 at 23:40
    
@Thomas, see revised version. –  Did Nov 15 '11 at 23:40
    
Thank you for your answer, sir! –  Robert Badea Nov 23 '11 at 7:25
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