Power Of 2

[MM]

Prove that
[tex]\left(36a+b\right)\left(36b+a\right)[/tex]
is not a power of two for any natural numbers a and b.

[masterfromkardjali]

Hallo,

I define

[tex]a = 2^{q}.p;[/tex]
[tex]b = 2^{f}.r;[/tex]

W.L.O.G.

[tex]q\ge f =>[/tex]

[tex]36.a +b = 36.2^q.p + 2^f.r = 2^f(36.2^{f-q} + r)[/tex]

If the assumption, that
[tex](36a + b).(36b + a);[/tex]
is a power of two, is correct [tex]=>[/tex]

[tex](36a + b);[/tex]
[tex](36b + a);[/tex]

are the both powers of two (we can prove that with induction). [tex]=>[/tex]

[tex](36a + b).(36b + a) = 2^f(36.2^{q-f}. p+r).(36b+a)[/tex]

obviously the odd factor

[tex](36.2^{q-f}. p+r)[/tex] is not power of two. Contradictio! q.e.d!

This problem was last semester my math homework, therefore I used my own notices. We can prove that also with the binary representations (algorithm for the powers of two).



Thank you for the attention,
masterfromkardjali