let $a,b,c\in \mathbb{N} $ and Quadratic equations $x^2-2ax+b=0,x^2-2bx+c=0,x^2-2cx+a=0$ has positive integer roots. find $a,b,c$

Question

let $a,b,c\in \mathbb{N} $ and Quadratic equations $x^2-2ax+b=0,x^2-2bx+c=0,x^2-2cx+a=0\,\,$ has positive integer roots. Find $a,b,c$

My working: WLOG $a\ge b\ge c$ and $a^2-b={\lambda}^2,\,\,b^2-c=\mu^2, \,\,c^2-a=\gamma^2; \lambda,\mu,\gamma\in \mathbb{N}$

Answer 1

Hint. Subtract the quadratic equations from each other. You get three linear equations in $x.$ Eliminate this from the system. Now you have a system for $a,b,c.$

Answer 2

This is what I got.

Consider the discriminants of the quadratic equations. These are:

$4a^2-4b$

$4b^2-4c$

$4c^2-4a$

Or you know what, let's just use the quadratic formula to lay down the roots of $x$ from each equation.

$(1) \ldots x=\dfrac{2a \pm \sqrt{4a^2-4b}}{2}=a \pm \sqrt {a^2-b}$

Similarly for the rest:

$(2) \ldots x=b \pm \sqrt {b^2-c}$

$(3) \ldots x=c \pm \sqrt {c^2-a}$

So here's what we know. Those constant terms are positive. Furthermore, in order for us to talk about integer solutions at all, the discriminants must be $0$ or perfect squares. Or at least the reduced expressions in the $\sqrt{}$

$\text{Case 1: Discriminants are null}$

$a^2-b=0 \implies b=a^2$

$b^2-c=0 \implies c=b^2=a^4$

$c^2-a=0 \implies a^8-a=0$

$a(a^7-1)=0 \implies a=\{0,1\}$

So either everything is $0$ and the equations don't exist or everything is $1$ and at least we have something. Hence here's a solution:

$\{a,b,c\}=\{1,1,1\}$

$\text{Case 2: Discriminants are slightly bigger}$

In all honesty, I'm stuck here. I'll update if I can get any more insight. But I guess that's half the work anyway.

Answer 3

The equations above $(1), (2), (3)$ are consecutive. From equation $(1)$ $x_{1,2}=\dfrac{2a\pm\sqrt{4a^2-4b}}{2}=a\pm\sqrt{a^2-b}\in\mathbb{ N}$. Then it should be $a^2-b=k^2\geq 0$. Given $a, b$ natural numbers. and $a^2-b=k^2<a^2$. Then it should be $a^2-b\leq (a-1)^2\Longleftrightarrow 2a-1\leq b.$ Analogous, we get $2b-1\leq c$ and $2c-1\leq a$. From the three it is obtained that $2(2(2a-1)-1)-1\leq 2(2b-1)-1\leq 2c-1 \leq a$. Then $8a-7\leq a\Longleftrightarrow a\leq 1$. It must be $a=1$ which results in $b=c=1$. So $(a, b, c)=\ \boxed{(1, 1, 1)}$