All pairs integers (m, n) such that 2^(m+1) + 3^(n+1) is perfect square

Question

zzr0ck3r · Answer

what is the question?

anonymous · Answer

Find all order pairs (m,n) which satisfy that eq

anonymous · Answer

2^(m+1) + 3^(n+1) = k^2

anonymous · Answer

By using trials i got one point (3,1). But i dont know with other solutions

anonymous · Answer

What about you @zzr0ck3r

rational · Answer

*

anonymous · Answer

is there a nice way, @rational :)

rational · Answer

not getting ideas sry
@dan815 @Kainui

rational · Answer

here is my attempt assuming $m,n$ are positive : 

$$2^{m+1}+3^{n+1}=k^2$$

Notice that $2^{m+1}\equiv 0\pmod{4}$ and $3^{n+1}\equiv (-1)^{n+1}\pmod{4}$.

Recall the fact that any perfect square is congruent to 0,1 in mod 4, therefore $n+1$ must be even. let $n+1=2r$ : 
$$2^{m+1}+3^{2r}=k^2$$

rearrange the equation and get
$$2^{m+1} = k^2-3^{2r}$$

factoring the right hand side
$$2^{m+1} = (k+3^r)(k-3^r)$$

rational · Answer

http://math.stackexchange.com/questions/1269526/find-all-pairs-of-positive-integers-m-n-such-that-2m13n1-is-a-per#1269552

kainui · Answer

I was trying but not able to get it to come together by relating the fact that every square is the sum of consecutive odd integers and that two consecutive triangular numbers also make a square with the binomial theorem and it written in a weird kind of way: $$\left( \frac{5+1}{2}\right)^{n+1} + \left( \frac{5-1}{2}\right)^{m+1} = k^2$$ I'll have to come back and think about this, interesting though!