I need to prove that $2^x$ never equals $3^y$, can anyone help?
Thanks
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6$\begingroup$ One is even, the other is odd? $\endgroup$– Ryan BudneyDec 4, 2013 at 12:59
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$\begingroup$ What can you assume? $\endgroup$– Tim RatiganDec 4, 2013 at 13:07
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$\begingroup$ @user113387 Are $x$ and $y$ integers? If they are not, then the proofs below don't work. In fact, if they are not, the result is not true (why?) $\endgroup$– LASVDec 4, 2013 at 13:32
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4 Answers
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According to the fundamental theorem of arithmetic every number $x \in\Bbb Z$ has a unique representation of product of prime numbers.
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$\begingroup$ en.wikipedia.org/wiki/Fundamental_theorem_of_arithmetic $\endgroup$ Dec 4, 2013 at 13:28
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That's not true. $x=y=0$ is a trivial solution.
Otherwise, the other comments seem to have good answers for $x,y>0$.
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Because those two numbers have different factors. One is composed of a bunch of 2s. The other is made up of 3s. How could they be the same?
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Ryan's comment has it right; the idea behind this problem is presumably to prove it without deferring to the fundamental theorem of arithmetic. Prove that every proper power of $2$, $2^x$ with $x>0$, is even, and that every proper power of $3$, $3^y$ with $y>0$, is odd. Since an odd number cannot equal an even number, you're done.
answered Dec 4, 2013 at 13:04