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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!

3 Answers 3

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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.

  • 0
    so no solutions?2012-07-19
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$\begin{eqnarray}{\bf Hint}\quad\ \rm mod\ 7\!:\ \ 2^n &\equiv&\rm\, -(a\!-\!b)^2 &\equiv&\rm\, (6a\!+\!b)(a\!+\!6b) \\ \rm\Rightarrow \ 1 \equiv 2^{3n} &\equiv&\rm\, -(a\!-\!b)^6 &\equiv&\rm\, -1\ or\ 0\ \ \Rightarrow\Leftarrow \end{eqnarray}$

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)/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)2012-07-12