Articles of vieta jumping

Resource for Vieta root jumping

I can’t seem to find a good resource on Vieta’s root jumping, which would explain various scenarios that suggest using the technique. Does anyone have a suggestion for a reference?

Positive integer solutions to $x^2+y^2+x+y+1=xyz$

The question asks for positive integer solutions to $x^2+y^2+x+y+1=xyz$ . We at first note that $x|y^2+y+1$. Now,let there exist positive integers $x,y$ that satisfy the given equation.Then $mx=y^2+y+1$ for some positive integer $m$ .Substituting this in the original equation yields $$x^2+x+mx=xyz$$ $$\implies x+1+m=yz$$ $$\implies \dfrac{y^2+y+1}{m}+1+m=yz$$ $$\implies y^2+y+m^2+m+1=myz$$ This equation’s exactly like the equation we were […]

Showing that $m^2-n^2+1$ is a square

Prove that if $m,n$ are odd integers such that $m^2-n^2+1$ divides $n^2-1$ then $m^2-n^2+1$ is a square number. I know that a solution can be obtained from Vieta jumping, but it seems very different to any Vieta jumping problem I’ve seen. To start, I chose $m=2a+1$ and and $n=2b+1$ which yields: $$ 4ka^2+4ka-4kb^2-4kb+k = b^2+b$$ […]

Math olympiad 1988 problem 6: canonical solution (2) without Vieta jumping

There is a recent question about this famous problem from 1988 on this forum, but I’m unable to respond to this because the subject is closed for me (insufficient reputation). Therefore this new post on the subject. here’s the link to the earlier subject The problem: Let a and b be positive integers. Let $$ […]

Diophantine quartic equation in four variables, part deux

A recent Question asked for all positive integer solutions of a simple quartic in four unknowns: $$ wxyz = (w+x+y+z)^2 \tag{1}$$ whose satisfaction is necessary for the integer side lengths $a,b,c,d$ of a cyclic quadrilateral having area equal to perimeter, where: $$ a+b+c+d = 2s = w+x+y+z $$ $$ w = s-a, x = s-b, […]

Integer solutions to the equation $a_1^2+\cdots +a_n^2=a_1\cdots a_n$

What is the general solution to the equation $$\sum_{j=1}^n a_j^2=\prod_{j=1}^n a_j,$$ $n\in \mathbb N$ , $n \ge 2$ over $\mathbb N_0$ ? WLOG, we can assume $0\le a_1 \le a_2\le \cdots \le a_n$ For every $n$, there is a solution, the trivial one: $a_1=\cdots =a_n=0$. For $n=2$ , the equation has only the trivial solution. […]

Is it true that $f(x,y)=\frac{x^2+y^2}{xy-t}$ has only finitely many distinct positive integer values with $x$, $y$ positive integers?

Prove or disprove that if $t$ is a positive integer, $$f(x,y)=\dfrac{x^2+y^2}{xy-t},$$ then $f(x,y)$ has only finitely many distinct positive integer values with $x,y$ positive integers. In other words, there exist $k\in\mathbb N$ such that if $n\gt k$ then $f(x,y)=n$ has no positive integer solutions. This problem is a generalization of this famous problem. Below is […]

Vieta Jumping: Related to IMO problem 6, 1988: If $ab + 1$ divides $a^2 + b^2$ then $ab + 1$ cannot be a perfect square.

The famous IMO problem 6 states that if $a,b$ are positive integers, such that $ab + 1$ divides $a^2 + b^2$, then $\frac{ a^2 + b^2}{ab + 1 }$ is a perfect square, namely, $gcd(a,b)^2$. How about a modification of this problem: If $a,b$ are (strictly) positive integers, such that $ab + 1$ divides $a^2 […]

Integer solutions to $\prod\limits_{i=1}^{n}x_i=\sum\limits_{i=1}^{n}x_i^2$

Given integers $x_1,\dots,x_n>1$. Let’s assume WLOG that ${x_1}\leq\ldots\leq{x_n}$. I want to prove that the only integer solutions to any equation of this type are: $x_{1,2,3 }=3\implies\prod\limits_{i=1}^{3}x_i=\sum\limits_{i=1}^{3}x_i^2=27$ $x_{1,2,3,4}=2\implies\prod\limits_{i=1}^{4}x_i=\sum\limits_{i=1}^{4}x_i^2=16$ I have a partial proof below, but there are a few holes in it that I would be happy to get help with. For $n=1$: $x_1<x_1^2$ $\color{green}{\text{Done.}}$ For […]

IMO 1988, problem 6

In 1988, IMO presented a problem, to prove that k must be a square if $a^2+b^2=k(1+ab)$, for positive integers a and b and k. I am wondering about the solutions, not obvious from the proof. Beside the trivial solutions a or $b=0$ or 1 with $k=0$ or 1, an obvious solution is $a=b^3$ so that […]