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Syler

#1 h May 29, 2010, 8:35 AM • 1 Y

Y by Adventure10

Let $a,b,k$ be positive integers such that
$k=\frac{a^2+b^2}{ab-1}$
Prove that $k=5.$

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jgnr

#2 h May 29, 2010, 11:38 AM • 3 Y

Y by sunny2000, Adventure10, Mango247

Suppose there exist $(a,b,k)\in\mathbb{N}^3$ such that $k=\frac{a^2+b^2}{ab-1}$ and $k\ne5$ . Let $S=\left\{(m,n)\in\mathbb{N}^2|\frac{m^2+n^2}{mn-1}=k\right\}$ . Take $(m,n)\in S$ such that $m+n$ is minimum, suppose without loss of generality that $m\ge n$ . If $m=n$ then $\frac{m^2+n^2}{mn-1}=\frac{2m^2}{m^2-1}=2+\frac2{m^2-1}$ , so $m^2-1\in\{1,2\}$ which is impossible. If $m=n+1$ then $\frac{m^2+n^2}{mn-1}=\frac{2n^2+2n+1}{n^2+n-1}=2+\frac3{n^2+n-1}$ , so $n^2+n-1\in\{1,3\}$ and $n=1$ which gives $k=\frac{2^2+1^2}{2\cdot1-1}=5$ , a contradiction. Therefore $m\ge n+2$ . Note that $\frac{t^2+n^2}{tn-1}=k$ is equivalent to $t^2-tkn+n^2+k=0$ . One of its roots is $t_1=m$ . The other root is $t_2=kn-t_1=\frac{n^2+k}{t_1}$ and hence is a positive integer, so $(t_2,n)\in S$ . By our assumption of the minimality of $m+n$ , we must have $t_2+n\ge m+n$ . So $\frac{n^2+k}m\ge m$ , $n^2+\frac{m^2+n^2}{mn-1}\ge m^2$ , $mn^3+m^2\ge m^3n-m^2$ , $2m\ge m^2n-n^3=n(m^2-n^2)\ge n(m^2-(m-2)^2)\ge4m-4$ , thus $m\le2$ which is impossible since $n\le m-2$ . Thus our assumption was wrong, therefore if $a,b,\frac{a^2+b^2}{ab-1}\in\mathbb{N}$ then $\frac{a^2+b^2}{ab-1}=5$ .

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#3 h May 29, 2010, 12:10 PM • 2 Y

Y by Adventure10, Mango247

Here is another way to solve it.

Consider a pair of $(a,b)$ satisfying the conditions.
Without loss of generality we may assume that there is a solution $a>b>1$ .
(Since we know $a=b$ has no solutions.)
Then we have $a^2-kab+(b^2+k)=0$ .

So there is another root called $a'$ that satisfies:
$a+a'=kb -(1)$
$aa'=b^2+k -(2)$
We know that $a'$ is also an integer (by $(1)$ ),
and $a'$ is positive (by $(2)$ ).

Hence $(a',b)$ is also a pair of solution.
But we have $a'-1=\frac{(b^2+k)-kb+1}{a-1}\leq \frac{b^2-3b+4}{b}\leq b-1$

The last inequality holds because obviously $k>2$ , and we assume that $b>1$ .
We jump from $(a,b)$ to $(b,a')$ .
Repeat these steps, and the loop will finally stop until one of them is $1$ .

Now take $b=1$ , then we get $a-1|a^2+1$ which leads to $a=2,3$ ,
both of them concludes $k=5$ .

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ArefS

#4 h Apr 4, 2011, 4:22 PM • 3 Y

Y by Amir Hossein, Adventure10, Mango247

here is my solution to this problem:
WLOG assume that $b\ge a$
notice that the hypothesis implies:
$ab-1\mid a^2+b^2$
$\implies ab-1\mid a^3+ab^2$ and we have: $ab-1\mid ab^2-b$ so $ab-1\mid a^3+b$ and therefore $\frac{a^3+b}{ab-1}$ is a positive integer.
We use strong induction on $ab$ .
we claim that there is a positive integer $c$ such that $\frac{a^2+c^2}{ac-1}=\frac{a^2+b^2}{ab-1}=k$ and $c<b$ .
assume that for a real $c$ we have that: $\frac{a^2+c^2}{ac-1}=\frac{a^2+b^2}{ab-1}=k$
$\iff (b-c)(a^3+b+c-abc)=0$
$\iff a^3+b=c(ab-1)$
$\iff c=\frac{a^3+b}{ab-1}$
so $c\in \mathbb N$
we have to prove that $c<b$ .
$\iff a^3<b(ab-2)$
we know that $b>a$ we only need to prove that $a^2<ab-2 \iff a(b-a)>2$
we consider 4 Cases:

$\bullet a=b$
Click to reveal hidden text

$\bullet a=b-1$
Click to reveal hidden text

$\bullet a=b-2$
Click to reveal hidden text

$\bullet b-a>2$
in the last case we obviously have $c<b$ so $ac<ab$ and $\frac{a^2+c^2}{ac-1}=\frac{a^2+b^2}{ab-1}=k$ .
According to the induction hypothesis we have that $\frac{a^2+c^2}{ac-1}=5$ and so $\frac{a^2+b^2}{ab-1}=5$ .

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