Vieta Root Jumping

Essentially, Mysterious 100 Degree Monic Polynomial shows we can even Vieta Jump for polynomials.

Another Exercise: Assume that $m$ and $n$ are odd integers such that

$m^{2} - n^{2} + 1 \mid n^{2} - 1$.

Prove that $m^{2} - n^{2} + 1$ is a perfect square.

Yet Another Exercise: Determine all pairs of positive integers $(a,b)$ such that

$\dfrac{a^2}{2ab^2-b^3+1}$

is a positive integer.

This one is one of my favorites!

Start out with noting that because $gcd(b, 4ab-1)=1$, we have: $4ab-1|(4a^2-1)^2$ $\iff 4ab-1|b^2(4a^2-1)^2$ $\implies 4ab-1 | 16a^4b^2-8a^2b^2+b^2$ $\implies 4ab-1 | (16a^2b^2)(a^2)-(4ab)(2ab)+b^2$ $\implies 4ab-1 | (1)(a^2)-(1)(2ab)+b^2$ $\implies 4ab-1|(a-b)^2$ The last step follows from $16a^2b^2\equiv (4ab)^2\equiv 1\pmod{4ab-1}$ and $4ab\equiv 1\pmod{4ab-1}$.

Let $(a,b)=(a_1, b_1)$ be a solution to $4ab-1|(a-b)^2$ with $a_1>b_1$ contradicting $a=b$ where $a_1$ and $b_1$ are both positive integers. Assume $a_1+b_1$ has the smallest sum among all pairs $(a,b)$ with $a>b$ , and I will prove this is absurd. To do so, I prove that there exists another solution $(a,b)=(a_2, b_1)$ with a smaller sum. Set $k=\frac{(a-b_1)^2}{4ab_1-1}$ be an equation in $a$. Expanding this we arrive at $4ab_1k-k=a^2-2ab_1+b_1^2$ $\implies a^2-a(2b_1+4b_1k)+b_1^2+k = 0$ This equation has roots $a=a_1, a_2$ so we can now use Vieta’s on the equation to attempt to prove that $a_1>a_2$. First, we must prove $a_2$ is a positive integer. Notice that from $a_1+a_2=2b_1+4b_1k$ via Vieta’s hence $a_2$ is an integer. Assume that $a_2$ is negative or zero. If $a_2$ is zero or negative, then we would have $a_1^2-a_1(2b_1+4b_1k)+b_1^2+k=0 \ge b^2+k$ absurd. Therefore, $a_2$ is a positive integer and $(a_2, b_1)$ is another pair that contradicts $a=b$. Now, $a_1a_2=b_1^2+k$ from Vieta's. Therefore, $a_2=\frac{b^2+k}{a_1}$. We desire to show that $a_2<a_1$. $a_2< a_1$ $\iff \frac{b_1^2+k}{a_1}<a_1$ $\iff b_1^2+\frac{(a_1-b_1)^2}{4a_1b_1-1}<a_1^2$ $\iff \frac{(a_1-b_1)^2}{4a_1b_1-1} < (a_1-b_1)(a_1+b_1)$ [ \iff \frac{(a1-b1)}{4a1b1-1} < a1+b1 ] Notice that we can cancel $a_1-b_1$ from both sides because we assumed that $a_1>b_1$. The last inequality is true because $4a_1b_1-1>1$ henceforth we have arrived at the contradiction that $a_1+b_1>a_2+b_1$. Henceforth, it is impossible to have $a>b$ (our original assumption) and by similar logic it is impossible to have $b>a$ forcing $a=b$ $\Box$

We want to force out a nice factorisation from $4ab - 1 \mid (4a^2-1)^2$. A few hit and trials lead me to consider:

$b^2(4a^2-1)^2 - (4ab-1)(4a^3b - 2ab + a^2) = (a-b)^2$.

So now, assume that there indeed exists distinct $(a,b)$ that satisfy $4ab-1 \mid (a-b)^2$.

Lemma: Suppose $(a,b)$ is a distinct positive integer solution to $\frac{(a-b)^2}{4ab-1} = k$ where WLOG $a > b$. Then $(b, \frac{b^2+k}{a})$ is also a solution.

Proof: Remark first that $(a-b)^{2}=(4ab-1)k ⇔a^{2}-(2b+4kb)a+b^{2}+k=0$. By Vieta's Formula for Roots, $\frac{b^2+k}{a} = 2b + 4kb \in \mathbb{Z^{+}}$ whence $(b, \frac{b^2+k}{a} )$ is also a solution. □

However, we easily see that $\frac{b^2+k}{a} = b(2+4k) > b$ since by assumption $k$ is positive. This means that given a solution $(a,b)$ where $b$ is taken to be minimum, we can always vieta jump to a smaller solution, a clear contradiction. ■

Problem: Show that if $x, y, z$ are positive integers, then $(xy+1)(yz+1)(zx+1)$ is a perfect square if and only if $xy+1, yz+1, zx+1$ are all perfect squares.