Find $a,b \in \mathbb{Z}^+$ such that : $\frac{a^{2}-2}{ab+2}\in \mathbb{Z}$

Question

$1$. Find $a;b\in \mathbb{Z}^+$ such that : $\frac{a^{2}-2}{ab+2}\in \mathbb{Z}$

$2$. Find $m;n>1$ such that : $2^m+3^n=k^2$ $(k\in \mathbb{Z})$

Problem 1. I thought :

$\frac{a^{2}-2}{ab+2}\in \mathbb{Z}\Rightarrow a^{2}-2\vdots ab+2\Rightarrow a^{2}+ab\vdots ab+2\Rightarrow ab(a+b)\vdots ab+2\Rightarrow 2(a+b)\vdots ab+2$

But i don't know how to do next!!

Answer 1

1

Case 1 : $\frac{a^2-2}{ab+2} =0$ no solution

Case 2 : $\frac{a^2-2}{ab+2} >0 $ and $a < b$ : So $$ \frac{a^2-2}{ab+2} < 1$$

no solution.

Case 3 : $\frac{a^2-2}{ab+2} >0 $ and $a \geq b\ (a>1,\ a>b)$ : So $$ \frac{a^2-2}{ab+2} < \frac{a^2}{ab} = \frac{a}{b},\ ba^2-2b < a^2b +2a$$ So $$ ab\leq ab(a-b)< 2(a+b),\ (a-2)(b-2)< 4$$

Hence if $b \geq 3$, then $a\leq 5$. In this case by computation we have $(a,b)=(4,3)$ (There was a solution-finding of questioner)

Case 4 : $\frac{a^2-2}{ab+2} <0 $ : no solution.

Answer 2

$2^m+3^n=k^2$

$m$ must be even,otherwise remainder of $2^m$ divided by $3$ will be $2$ thus $k^2=3w+2$ that is impossible.

$m=2t$

$3^n=k^2-2^{2t}=(k-2^t)(k+2^t)$

So

$k-2^t=3^a,k+2^t=3^b$

Then we will have: $2.2^t=3^b-3^a \to 2^{t+1}=3^a(3^{b-a}-1)\to a=0,3^{b}-1=2^{t+1} \to b\in\{1,2\}$ So $t\in\{0,2\} \to k\in\{4,5\}\to (m,n)\in\{(0,1),(4,2)\}$

But $m,n>1$ so $(m,n)=4,2$.

To know why $3^b-1=2^{t+1}$ has only two answers, look at here.

Answer 3

2

We note that $(m,n) = (4,2)$ is a solution , we guess that it is the only one.Note that $3$ doesn't divide $k$, and hence $k^2 \equiv 1\pmod{3} $.

Case 1: If $m$ is odd we can deduce from Fermat's little theorem that $$ 1 \equiv k^2 \equiv 2^m + 3^n \equiv 2^{2s+1} +0 \equiv 2 \pmod{3} $$

, which is a contradiction.

Case 2: If m is even, we let $m=2s$ with $s > 0$, we can then write the equation as $$3^n = k^2 - 2^{2s} = (k - 2^s)(k + 2^s)$$, We note that $ \gcd ((k -2^s),(k + 2^s )) | 2^{s+1} $, so $3$ cannot divide both factors, thus if we have a solution we must have $k - 2^s = 1$ which implies $k = 2^s + 1 $, so it suffices to solve $$3^n - 1 = 2^{s+1}$$ (Which seems a little easier at least we have only two unknowns).

There is a known theorem (and technique) known as "Lifting exponent lemma" you can read this article , which say that (denoting $v_p(n)$ to be the largest exponent of $p$ dividing n ) if $a \equiv b \pmod{2}$ , and $v_2(\frac{a^2-b^2}{2}) = \alpha $ , $v_2(x)= \beta$ , then $v_2(a^x - b^x) = \alpha + \beta $ (in fact that theorem arises naturally in practice in special cases )

So in our equation in order to have a solution we must have $v_2(3^n-1)=s+1$ , but $v_2(\frac{3^2 - 1}{2}) = 2 $ , thus $v_2(n) = s-1 > 0 $ (since $n > 1$), We have

• If $n=2$, then $s=2$ is a solution , which corresponds to $m=4$

• If $n>2$, we must have $2^{s-1} \| n $ , if $s=2$ , then $s=2m$ , with $m > 1$ , but then it is clearly not solution. If $s > 2$ , then by induction one can prove that $$2^{s+1} < 3^{2^{s-1}} - 1 < 3^{2^{s-1}f} - 1 $$, when $f > 1$ .

So the only solution is that we have got.