I recently came across the "infamous" Question 6 after watching this youtube video by Numberphile. Just to refresh everyone's memory, here it is:
Let $a$ and $b$ be positive integers such that $ab + 1$ divides $a^2 + b^2$. Show that
$$\frac{a^2+b^2}{ab+1}$$ is the square of an integer.
Being ever the optimist, I took out a pen and paper and got to a very simple solution in less than 30 minutes. With all this hoo-hah about this problem, I assumed my solution is probably wrong, so I decided to upload here. As soon as I began to write it in a more orderly fashion, I realized I have a missing step :(
I still think this approach is very elegant (a lot more than those Vietta jumps which are basically a Deux Ex-Machina solution imo). I'm hoping someone here can fill in the missing piece?
My proof:
Let $c$ be the quotient which results from the division: $$\frac{a^2+b^2}{ab+1}=c$$ Moving the denominator to RHS we get: $$a^2+b^2=abc+c$$ Using the fundamental theorem of arithmetic we write every term in the canonical representation - as a unique product of powers of primes, where ${p_i}$ are the prime numbers: $$\left(\displaystyle\prod_{i=1}^{\infty} p_i^{a_i}\right)^2 +\left(\displaystyle\prod_{i=1}^{\infty} p_i^{b_i}\right)^2= \displaystyle\prod_{i=1}^{\infty} p_i^{a_i}\displaystyle\prod_{i=1}^{\infty} p_i^{b_i}\displaystyle\prod_{i=1}^{\infty} p_i^{c_i}+\displaystyle\prod_{i=1}^{\infty} p_i^{c_i} $$ It follows that: $$\displaystyle\prod_{i=1}^{\infty} p_i^{2a_i} +\displaystyle\prod_{i=1}^{\infty} p_i^{2b_i}= \displaystyle\prod_{i=1}^{\infty} p_i^{{a_i}+{b_i}+{c_i}}+\displaystyle\prod_{i=1}^{\infty} p_i^{c_i} $$We shall now show that for every $i$, $c_i$ is even
For every $i$ where $a_i \neq b_i$
We shall examine the p-adic order ${\nu_p}_i$ of both sides, where ${\nu_p}_i$ is defined to be the highest integer power such that the expression is divisible by ${p_i}^{{\nu_p}_i}$.
Note that at least one of $a_i,b_i$ in nonzero. From RHS we see that ${\nu_p}_i$ is exactly $c_i$ (if it were larger by 1, second term would become a fraction while left term remains an integer, so the sum would be a fraction)
From LHS by using the same exact consideration we see that ${\nu_p}_i$ is $\min(2a_i,2b_i)$.
In this case ${\nu_p}_i$ and hence $c_i$ is even.For every $i$ where $a_i = b_i$
This is where I'm stuck! I want to show that $c_i$ is even. So much trickier than I thought!From (1) and (2) we see that $c_i$ is even for every i, hence: $$\sqrt{c}=\sqrt{\displaystyle\prod_{i=1}^{\infty} p_i^{c_i}}=\displaystyle\prod_{i=1}^{\infty} p_i^{\frac{c_i}{2}}\in \mathbb{N}$$ QED???
Can someone help fill-in the gap?
Your approach is flawed from the very beginning. I assume your product is ranging over all possible prime numbers because of the infinite product notation. In that case, your claim that both $a_i$ and $b_i$ cannot be zero is trivially false.
$a,b,c$ are finite numbers, so for large enough prime $p_i>\max\{a,b,c\}$, the exponents are both zero.
If your infinite product notation is just an unfortunate mistake and you meant to use the correct canonical prime factorization, then you cannot assume that $a$, $b$ and $c$ all have same set of prime divisors.