# 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!

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Hint: $(6a+b)(a+6b)$ is a power of $2$ iff $(a+6b)$ and $(6a+b)$ are powers of $2$ individually.

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please elaborate –  fosho Jul 11 '12 at 19:44
@Daniel: The only factors of a power of $2$ are powers of $2$, so there must be integers $m,n$ such that $6a+b=2^m$ and $a+6b=2^n$. Without loss of generality assume that $m\le n$. Then $7(a+b)=(6a+b)+(a+6b)=2^m+2^n$. Now factor that last expression. –  Brian M. Scott Jul 11 '12 at 20:04
take out factor of 2? –  fosho Jul 11 '12 at 22:11
@Brian Out of curiosity, what method did you have in mind to finish that proof? –  Bill Dubuque Jul 12 '12 at 2:17
@Bill: $7\mid 2^{n-m}+1$, but powers of $2$ are congruent to $1,2$, or $4$ mod $7$, not $-1$. –  Brian M. Scott Jul 12 '12 at 2:28

$\begin{eqnarray}{\bf Hint}\quad\ \rm mod\ 7\!:\ \ 2^n &\equiv&\rm\, -(a\!-\!b)^2 &\equiv&\rm\, (6a\!+\!b)(a\!+\!6b) \\ \rm\Rightarrow \ 1 \equiv 2^{3n} &\equiv&\rm\, -(a\!-\!b)^6 &\equiv&\rm\, -1\ or\ 0\ \ \Rightarrow\Leftarrow \end{eqnarray}$

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 $\rm\:2^n\equiv 0\:\Rightarrow\:7\:|\:2.$

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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.

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(2^n)/7 will always return one of the value from the SET (2,1,4,9). Therefore (a + b) will always be one of the value from the SET (3,6,11,5,10,13) –  Rajesh K Singh Jul 12 '12 at 3:49
(2^n)/5 will always return one of the value from the SET (2,4,3,1). Therefore (a - b) will always be one of the value from the SET (-3,-2,-1,0,1,2,3) –  Rajesh K Singh Jul 12 '12 at 3:55