# A claim that a is a square

We have integers $a,b,c,d$ such that $a<b\le c<d$ and $ad=bc$ and $\sqrt{d}-\sqrt{a}\le 1$.Show that $a$ is a perfect square.This question is pretty unbelievable for me.anyway I don't know if I am reposting this here.

• @panoramix: With $\sqrt d - \sqrt a = 1$ we have infinitely many solutions: $a = k^2$, $d=(k+1)^2$, $b, c = k(k+1)$ for all $k \in \Bbb{N}$. – Daniel R Jan 9 '14 at 15:01

Let us show that $a = n^2 ,\; b = n^2+n = c,\; d = (n+1)^2$ is the only solution.
The condition $\sqrt{d}-\sqrt{a} \leq 1$ shows that $$d\leq a+2\sqrt{a}+1.$$
Let $b = a+x,\; c = a+y,\; d = a+z$ with $x,y,z \in \Bbb N$. Then $0<x\leq y < z \leq 2\sqrt{a}+1$, showing $0<x\leq y \leq 2 \sqrt{a}$. Furthermore:
$$ad = bc = (a+x)(a+y) = a^2+(x+y)a+xy$$ and reducing $\bmod a$ shows $a \mid xy$. Thus $$a\leq xy$$ and there is $k \in \Bbb N$ such that $xy = ak$. This leads to $ad = bc= (a+x)(a+y) = a(a+x+y+k)$ and dividing by $a$ yields $a+x+y+k = d \leq a+2\sqrt{a}+1$. As $k\geq 1$, we conclude $$x+y\leq 2\sqrt{a}.$$
We find $4a \leq 4xy \leq 4xy +(x-y)^2 = (x+y)^2 \leq 4a$. Hence $x=y$ by $(x-y)^2 = 0$. Finally $x+y = 2\sqrt{a}$ implies that $a$ is necessarily a perfect square.
• I assume that $a \geq 0$ which implies $1\leq a<b\leq c<d$. – benh Jan 9 '14 at 16:40