Question

Let a, b, c be positive real numbers. The sequence $\left(a_n\right)n\ge 1$ is defined by$a_1=a,a_2=b$ and$a_{n+1}=\frac{a_n^2+c}{a_{n-1}}$ for all $n\ge 2.$ Prove that the terms of the sequence are all positive integers if and only if a,b and $\frac{a^2+b^2+c}{ab}$ are positive integers.

• A. $c=a_s{a}_s+2-a_s^2$
• B. $c=a_s{a}_s-2-a_s^2$
• C. $c=a_s{a}_s-2+a_s^2$
• D. $c=a_s{a}_s+2+a_s^2$

Answer 1

The correct option is A $c=a_s{a}_s+2-a_s^2$

Clearly, all the terms of the sequence are positive numbers. Write the recursive relation as $a_n{+}_1{a}_{n-1}=a_n^2+c.$
$a_n{a}_{n-2}=a_{n-1}^2+c,$
and by subtracting the two equalities we deduce $a_{n-1}(a_{n+1}+a_{n-1})=a_n(a_n+a_{n-2}).$
Therefore $\frac{a_{n+1}+a_{n-1}}{a_n}=\frac{a_n+a_{n-2}}{a_{n-1}}$
$n\ge 3$ for all
$b_n=\frac{a_{n+1}+a_{n-1}}{a_n}$ It follows that the sequence
$b_n=k,$ is constant, say
For all $n\ge 2.$ Then the sequence $\left(a_n\right)$ satisfies the recursive relation $a_{n+1}=k{a}_n-a_{n-1}$ for all $n\ge 2$ and since $a_3=\frac{b^2+c}{a}=kb-a,$
We derive that $k=\frac{a^2+b^2+c}{ab}$ Now, if a,b and k are positive integers, it follows inductively that $a_n$
is a positive integer for all $n\ge 1$
Then a,b are positive integers and $k=\frac{a_3+a}{b}$ is a rational number .Let
$k=\frac{p}{q},$
where p and q are relatively prime positive integers. We want to prove that q = 1. Suppose that $q>1.$
From the recursive relation we obtain $q(a_{n+1}+a_{n-1})=p{a}_n,$
and hence q divides $a_n$ for all $n\ge 2$ We prove by induction on s that
$q^s$ divides $a_n$ for all $n\ge s+1$ We have seen that this is true for s=1. Suppose $q^{s-1}$ divides
$a_n$ for all $n\ge s.$ we have $a_{n+2}=\frac{p}{q}{a}_{n+1}-a_n$,which is equivalent to$\frac{a^{n+2}}{q^{s-1}}=p\frac{a_{n+1}}{q^s}-\frac{a_n}{q^{s-1}}.$ if $n\ge s,$ then,
$q^{s-1}$ divides $a_n$ hence $a_n+2,$ and $q^s$ divides $a_{n+1}.$ it follows that $q^s$
$a_{s+2}=\frac{a_{s+1}^2+c}{a_s}$
$c=a_s{a}_s+2-a_s^2$
which implies that c is divisible by $q^{2(s-1)}$ for all
$s\ge 1.$ because $c>0,$ this is a contradiction.