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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
$(m+2n)\ln n = 4n \ln m$ 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.

share|cite|improve this answer
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

Your Answer


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.