Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question

2 Answers 2

up vote 3 down vote accepted

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 $x$.

share|improve this answer
    
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

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

...

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

share|improve this answer
3  
I would like to see an elaboration of your idea then. –  Karl Kronenfeld Jul 24 '13 at 1:23

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.