When is $(6a + b)(a + 6b)$ a power of two? 
Find all positive integers $a$ and $b$ for which the product $(6a + b)(a + 6b)$ is a power of $2$.

I havnt been able to get this one yet, found it online, not homework!
any help is appreciated thanks!
 A: $\begin{eqnarray}{\bf Hint}\quad\ \rm mod\ 7\!:\ \ 2^n &\equiv&\rm\, -(a\!-\!b)^2 &\equiv&\rm\, (6a\!+\!b)(a\!+\!6b) \\
\rm\Rightarrow \ 1 \equiv 2^{3n} &\equiv&\rm\, -(a\!-\!b)^6 &\equiv&\rm\, -1\ or\ 0\ \ \Rightarrow\Leftarrow
\end{eqnarray}$
Remark $\ $ Said conceptually, if it were solvable  then $-1$ would be a square mod $7,\: $ since we have $\,2\equiv 3^2\,$ so $\rm\:2^n\equiv 3^{2n}\equiv -(a\!-\!b)^2\:\Rightarrow\:   (3^n/(a\!-\!b))^2\equiv -1.\:$ Note $\rm\:a\not\equiv b\:$ else $\rm\:2^n\equiv 0\:\Rightarrow\:7\:|\:2.$
A: Hint: $(6a+b)(a+6b)$ is a power of $2$ iff $(a+6b)$ and $(6a+b)$ are powers of $2$ individually. 
A: Assume, (6a + b)(a + 6b) = 2^c
where, c is an integer ge 0
Assume, (6a + b) = 2^r
where, r is an integer ge 0
Assume, (a + 6b) = 2^s
where, s is an integer ge 0
Now, (2^r)(2^s) = 2^c
i.e. r + s = c
Now, (6a + b) + (a + 6b) = 2^r + 2^s
i.e. 7(a + b) = 2^r + 2^s
i.e. a + b = (2^r)/7 + (2^s)/7
Now, (6a + b) - ( a + 6b) = 2^r - 2^s
i.e. a - b = (2^r)/5 - (2^s)/5
Now we have,
a + b = (2^r)/7 + (2^s)/7
a - b = (2^r)/5 - (2^s)/5
Here solving for a we get, a = ((2^r)(6) - 2^s)/35
Now solving for b we get, b = ((2^s)(6) - 2^r)/35
A careful observation reveals that there are no positive integers a and b for which the product (6a + b)(a + 6b) is a power of 2.
