Let $x, a, b$ natural numbers such that $x^{a+b}=a^b b$. How prove: $a=x$ and $b=x^x$?
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2What have you ytied? – Thomas Andrews Jul 24 '14 at 12:04
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1$ln(x^a \cdot x^b)=ln(a^bb)$ and I do not know what to do next – piteer Jul 24 '14 at 12:10
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2Are you struggling to prove that $a=x, b=x^x$ satisfy the equation, or do you want to prove that these are the only satisfying values? – Karolis Juodelė Jul 24 '14 at 12:11
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Original form: Let $x, a, b$ natural numbers such that $x^{a+b}=a^b b$.Prove that $a=x$ and $b=x^x$. – piteer Jul 24 '14 at 12:15
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a,b,x natural numbers – piteer Jul 24 '14 at 12:21
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2Probably not helpful to use logarithms, because this is a problem about natural numbers. Think in terms of unique factorization of these numbers. – Thomas Andrews Jul 24 '14 at 12:35
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ok, but do not know how? – piteer Jul 24 '14 at 12:37
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Let $p$ be any prime and let $v$ denote the corresponding valuation (i.e., $v(n)=k$ if $p^k\mid n$ and $p^{k+1}\nmid n$). Then $$\tag1(a+b)v(x)=bv(a)+v(b). $$ If $v(a)<v(x)$, we find $v(b)=av(x)+b(v(x)-v(a))\ge a+b\ge b>v(b)$, contradiction. Therefore $v(a)\ge v(x)$. Say, $v(a)=v(x)+\delta$ with $\delta>0$. Then $(1)$ becomes $$ av(a) +(a+b)\delta=v(b)$$ If $\delta>0$, we again arrive at a contradiction: $$v(b)=av(a)+(a+b)\delta\ge b\delta\ge b>v(b).$$ We conclude $\delta=0$, i.e. $v(a)=v(x)$. As $p$ was arbitrary (and $a,x$ are positive), we conclude $a=x$, and then clearly $b=x^x$.
Hagen von Eitzen
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There seems to be an error in this proof: where you have $v(b) = av(a) + (a+b)\delta$, I get $v(b) = av(a) - (a+b)\delta$. The result is that your chain of inequalities doesn't work, and I don't see how to fix it. – mathmandan Feb 10 '15 at 16:19