13
$\begingroup$

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.

$\endgroup$
1
  • 3
    $\begingroup$ @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 $k \in \Bbb{N}$. $\endgroup$ – Daniel R Jan 9 '14 at 15:01
10
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ I assume that $a \geq 0$ which implies $1\leq a<b\leq c<d$. $\endgroup$ – benh Jan 9 '14 at 16:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.