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I will be very grateful, if someone show me, how to solve such equations.

Example 1.

$$ n^{m+2n}=m^{4n} $$

n,m - positive integers. Thanks a lot.

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$(m+2n)\ln n = 4n \ln m$ implies $\log_n(m)=\frac{m+2n}{4n}$ is rational, hence $n=k^r$, $m=k^s$ with $k,r,s$ positive integers. But then we have to solve $(k^s+2k^r)\cdot r = 4sk^r$ – Hagen von Eitzen Sep 10 '12 at 9:41
$n^m=(\frac{m^2}{n})^{2n}\implies n\mid m^2$, also either $n$ is perfect square or $m$ is even – lab bhattacharjee Sep 10 '12 at 9:47
Taking Hagen's comments we can rearrange to give $rk^s=2(2s-r)k^r$ – Mark Bennet Sep 10 '12 at 9:50
Note also the trivial solution $m=n=1$ – Mark Bennet Sep 10 '12 at 10:03
If prime $p\mid n,p\mid m$, SO if $n=\prod p_i^{q_i}, m$ must be of the form $\prod p_i^{r_i}$ – lab bhattacharjee Sep 10 '12 at 10:44
up vote 1 down vote accepted

My personal choice of dealing with such equations is to do the following: Take any prime dividing $m$. It is straightforward to see that it must also divide $n$. Conversely, if you take any prime dividing $n$, it must also divide $m$. Hence, $m,n$ have the same primes dividing them.

Now, take any prime $p$ dividing them both. Define $v_p(x)$ to be the highest exponent of $p$ in $x$. As $m^{4n} = n^{m+2n}$, therefore, $ (m+2n) v_p(n) = (4n) v_p(m)$.

Now, comes the main step. Notice now, that if $(m+2n)> 4n$, then $v_p(n) < v_p(m)$, and it is true for all primes dividing $m,n$, hence $n|m$. On the other hand, if $m+2n < 4n$, then $v_p(m) <v_p(n)$, and thus $m|n$. Finally, if $m+2n = 4n$, then $n=m$.

The rest is simple case analysis. See that $m+2n = 4n \implies m=2n$, a contradiction with $m=n$, hence the last case isn't possible.

Next, if $m+2n < 4n$, then $m|n$ implies $n = km$, for some positive $k$. Hence, $$(2k+1)m( v_p(k) + v_p(m)) = 4km v_p(m) \implies (2k+1) v_p(k) = (2k-1) v_p(m)$$ Because $2k-1,2k+1$ are coprime, therefore, $2k-1|v_p(k)$. Therefore, $k \ge p^{2k-1} \ge 2^{2k-1} > k$ for $k>1$. Hence, $k=1$, and this yields the solution $m = n=1$.

Finally, if $4n < m+2n \implies m > 2n$, then $n|m$ implies $m = kn$ for $k \ge 3$. Hence, $$(k+2)n v_p(n) = 4n v_p(n) + 4n v_p(k) \implies (k-2)v_p(n) = 4 v_p(k)$$ Now, if $k$ is odd, then $k-2 | v_p(k)$, but $3^{k-2} > k$ for $k \ge 4$. If $k=3$, then $n=3^4, m=3^5$, which is a valid solution.

If $k$ is even, then if $k=4t$ for $t \ge 1$, then $(2t - 1) v_2(n) = 2(2+ v_2(t))$, hence, $2t-1 | 2 + v_2(t) \implies t \ge 2^{2t-3} $ which is not true for $t \ge 3$. If $t=1$, then $k=4$, and $n = 16$ (From $(k-2)v_p(n) = 4 v_p(k)$), so the solution is $(16,64)$. If $t=2$, then $k=8$, and $n=4$ leading to $(4,32)$ being a solution.

Finally, if $k=2t$, where $t$ is odd, then $\frac{(t-1)}{2} v_p(n) = v_p(k)$. Note that $k \ge 3$ implies $t \ge 3$, as $t$ is odd. Choose a prime $q$ dividing $t$. Then, $\frac{(t-1)}{2} v_q(n) = v_q(k) = v_q(t)$, so $(t-1)/2 | v_q(t) \implies t \ge 3^{(t-1)/2} $ which doesn't hold for $t \ge 5$. Hence $t = 3, k=6$, and $n=6, m=36$, so $(6,36)$ is a solution.

Thus, the only solutions are $(1,1),(4,32),(6,36),(16,64),(81,243)$.

A similar question was asked in the IMO 1997 as Problem 5.

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Using Java program, $(4,32),(16,32)$ seem to be solutions. – lab bhattacharjee Sep 10 '12 at 11:44
Oh I missed the case when $k$ is even in the last case. Let me fix it. – Rijul Saini Sep 10 '12 at 11:53
@labbhattacharjee (16,32) isn't a solution, as $m = 2n \implies m=n$, by cancelling the powers. – Rijul Saini Sep 10 '12 at 12:27
Sorry, it is actually $(16,64)$ – lab bhattacharjee Sep 10 '12 at 13:07

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