The equation, $xk = x^k$ (where $x$ and $k$ are both integers).
Are there any solutions other than $\{ (1,1), (2,2) \}$ ?
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1$k = 1$ works for all $x$. – Daniel Fischer Oct 04 '13 at 16:10
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yeah, forgot to mention – tmj Oct 04 '13 at 16:10
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1This is equivalent to $x^{k-1}=k$ and $2^{k-1}>k$ for $k>2$. – njguliyev Oct 04 '13 at 16:12
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If not, just write $x=1+y$ and use the binomial theorem – Prahlad Vaidyanathan Oct 04 '13 at 16:12
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@PrahladVaidyanathan could you please elaborate. – tmj Oct 04 '13 at 16:13
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Oh, and $x = 0, , k > 0$. – Daniel Fischer Oct 04 '13 at 16:13
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Well, $x^k=kx$ implies $x(x^{k-1}-k)=0$. Thus, $x=0$ works for all $k>0$ and $k=1$ works for all $x$. Aside from that, there are only solutions when $k$ has a $(k-1)$-th root, which I believe only occurs when $k=1$ or $k=2$.
Clayton
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A $(k-1)$-th root gives the unique positive real number such that, when raised to the $k-1$ power, gives the number you are interested in. (Think of square root, cube root, etc.) For your question, when is the $(k-1)$-th root not just a real number, but an integer. This means if you look at all of the prime factors of the integer, the exponents of the primes are divisible by $k-1$. – Clayton Oct 04 '13 at 16:15
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There are no solutions unless $k=1$ or $x=k=2$. If $k=2$, $x^2=x*x \gt kx$ unless $x=2$. If $k \gt 2, 2^k \gt 2k$ as $2^k=(1+1)^k\gt 1+k+\frac 12k^2 \gt 2k$ and higher $x$ makes it worse.
Ross Millikan
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