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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
 elim Moderator Posts: 578 Joined: Feb 2010 Reputation: 0
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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