[Math] Prove that there exists infinitely many pairs of relatively prime integers $(a,b)$.

elementary-number-theory

Prove that there exists infinitely many pairs of relatively prime integers $(a,b)$ such that both the quadratic equations $$x^2+ax+b=0$$
$$
x^2+2ax+b=0$$
has integer roots.

I tried the following:
$a^2-4b=n^2$ and $a^2-b=m^2$

Eliminating $b$ from both we get

$3a^2=(2m+n)(2m-n)$

from here we get that $3a^2$ should have two factors whose sum is divisible by $4$. From here we can get the $b$ but the problem is both numbers should be relatively prime.

Best Answer

You want $$ 3a^2 - 4 m^2 = -n^2. $$ Suppose you have a single such expression, $a=1,m=1,n=1,$ call it $(a,m) = (1,1).$ You get infinitely many such expressions, with the same fixed value of $m,$ by repeating the mapping $$ (a,m) \mapsto (7a+8m, 6a+7m). $$ So, we have $(a,m)$ pairs $$ (1, 1) $$ $$ (15, 13) $$ $$ (209, 181) $$ $$ (2911, 2521) $$ $$ (40545, 35113) $$

We also get separate recursions for the $(a,m)$ components, $$ a_{j+2} = 14 a_{j+1} - a_j, $$ $$ m_{j+2} = 14 m_{j+1} - m_j. $$ This is just the Cayley-Hamilton theorem. Similarly, the mapping preserves GCD, because $$ \left( \begin{array}{rr} 7 & 8 \\ 6 & 7 \end{array} \right) $$ is of integers and determinant $1,$ that is, its inverse is also of integer entries. Back to Cayley-Hamilton, what is the characteristic polynomial for this matrix?

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