# If $m,n\in\mathbb{N}$ and $f(m)|f(n)$ for the $f$ given below, must $m=n$?

If for two natural number $m$ and $n$, $(2^{2m+1}-1)2^{4m-2}(2^{2m+1}+2^{m+1}+1)\mid(2^{2n+1}-1)2^{4n-2}(2^{2n+1}+2^{n+1}+1)$, then $m=n$?

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Another answer. Let $p(z)=(2z^2-1)(2z^2+2z+1)$. Then you want $m<n$ and $p(2^m)|p(2^n)$.

Given $m$, let $D_m=p(2^m)$. Since $D_m$ is odd, if $n\equiv m \pmod {\phi(D_m)}$, then $2^n \equiv 2^m \pmod {D_m}$ and thus $D_n=p(2^n)\equiv P(2^m)=D_m \equiv 0\pmod {D_m}$

So for any $m$, there are infinitely many $n$.

Actually, you can show that $$2^{8m+4}\equiv 1\pmod {2^{2m+1}+2^{m+1}+1}$$ and $$2^{8m+4}\equiv 1\pmod {2^{2m+1}-1}$$

So $2^{8m+4}=1\pmod {D_m}$. So, if $m\equiv n \pmod {8m+4}$, then $p(2^m)|p(2^n)$.

E.g., if $m=2$, $8m+4=20$, and we get $D_2|D_{22}$.

More generally, if $p(z)$ is an integer polynomial with $p(0)$ odd, then for any $m$, there are infinitely many $n>m$ such that $p(2^m)|p(2^n)$.

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+1 There you have it. A recipe for infinitely many counterexamples. – Jyrki Lahtonen Dec 24 '12 at 17:24
Yep, and your example fits, since $2^1\equiv 2^{13}\pmod {7\times 13}$, so $D_1|D_{13}$. – Thomas Andrews Dec 24 '12 at 17:27

False as stated. The divisibility also holds for example, when $n=1$ and $m=13$.

When $m=1$, the left hand side is $$(2^3-1)2^2(2^3+2^2+1)=7\cdot4\cdot13.$$ When $n=13$ we obviously have that $$(2^3-1)\mid(2^{27}-1)\qquad\text{and}\qquad2^2\mid2^{4\cdot13-2}.$$ The last factor is a bit trickier, but from Little Fermat we see that modulo $13$ we have the congruences $2^{13+1}\equiv2^{1+1}$ and $2^{2\cdot13+1}\equiv2^{2\cdot1+1}$, so also $$2^{27}+2^{14}+1\equiv2^3+2^2+1\equiv0\pmod{13}.$$ It seems to me that is should be easy to find several other examples of `factor wise' divisibility.

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