# an injection into $\mathbb{N}$

Is that true that the map $f\colon \{(m,n)\in\mathbb N^2:m\le n\}\to\mathbb N$ defined by $(m,n)\mapsto (m+n)^{\max\{m,n\}}$ is an injection? If it is, how to prove that? I have asked a similar question but it appeared to be very easy. My original struggling in the previous question was that I assumed that $m\leq n$ instead of checking easy reasons for the map not to be injective.

Thanks!

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Suppose $f(m,n)=f(m',n')$, i.e. $(m+n)^n=(m'+n')^{n'}$. If $n=n'$ we get $m'=m$, so assume without loss of generality that $n<n'$. If $m+n\le m'+n'$, then $$(m+n)^{\max(m,n)}=(m+n)^n< (m+n)^{n'}\le (m'+n')^{n'}=(m'+n')^{\max(m',n')}$$ so $f(m,n)\neq f(m',n')$. Hence we may conclude $m+n>m'+n'$.

Write $a=m+n, b=m'+n'$, with $a^{n}=b^{n'}$, an integer assumed bigger than 1. For any prime $p$ dividing both $a,b$ we have $n\nu_p(a)=n'\nu_p(b)$, so $\nu_p(a)>\nu_p(b)$. Hence $b|a$, and since they are unequal $a\ge 2b$ or $$m+n\ge 2(m'+n')$$

On the other hand, since $m\le n<n'$, we have $$(m+n)^n<(2(m'+n'))^n$$

Edit: $\nu_p(x)$ denotes the valuation, i.e. the maximum number of $p$'s that divide integer $x$.

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That's nice! Thanks a lot! – Sasha Patotski Jul 24 '13 at 2:52
You're welcome, glad to help. – vadim123 Jul 24 '13 at 2:57

$$(2n)^n-(n+2)^{(n+1)}>0.$$

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for $n$ large (starting from not so large).

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I would like to see an elaboration of your idea then. – Karl Kronenfeld Jul 24 '13 at 1:23