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Interesting: $\displaystyle{\frac{a^{2}+b^{2}}{ab+1}}$ is a perfect square if it's an integer (\(a,b \in \mathbb{N}^+\))
05-10-2010, 08:52 AM
Post: #1
Interesting: $\displaystyle{\frac{a^{2}+b^{2}}{ab+1}}$ is a perfect square if it's an integer (\(a,b \in \mathbb{N}^+\))
Suppose that $a,b,k=(a^{2}+b^{2})/(a b + 1) \in \mathbb{N}^+$, then there are \(a_0,b_0 \in \mathbb{N}^+\) such that
\(a_0+b_0\) is the smallest with \(k=(a_0^{2}+b_0^{2})(a_0 b_0 + 1)\)

If $a_0= b_0$, then $2a_0^2= k(a_0^2+1) \Rightarrow a_0=b_0=k=1 =1^2$.

Assume $a_0< b_0$. Since \((a_0^2+b_0^2)/(a_0 b_0+1) > (a_0^2+b_0^2)/b_0^2>1\) ,
we have \(k>1\) and \((ka_0)^2-4(a_0^2-k)>0\)

Let $b_1$ be the root of the equation $b^2-ka_0b+a_0^2-k=0$ other than \(b_0\),
we then have $b_1 \ne b_0$ and so $b_1+b_0-ka_0 = 0$

Because $k=(a_0^2+b_0^2)/(a_0b_0+1) > b_0/a_0-1/a_0$, $b_1=ka_0-b_0 \ge 0$, ie $b_1 \in \mathbb{N}$.

If $b_1>0$, then by minimum property of $b_0$, $b_1>b_0$ hence $ka_0=b_1+b_0>2b_0$,
and $(a_0^2+b_0^2)/(a_0b_0+1) = k\ge 2b_0/a_0$. Consequently, \(0<a_0(a_0^2+b_0^2)-2b_0(a_0b_0+1)= a_0^3-a_0b_0^2-2b_0<a_0(a_0+b_0)(a_0-b_0)\)
and so \(b_0 < a_0\). A contradiction.

Therefore \(b_1 = 0\) and \(k = a_0^2\) is a perfect square.
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