Does there exist $m,a\in \mathbb{N}$ such that $m(m+1)=a^7$? Let $m$ be a natural number. Can  $m(m+1)$ be written as a seventh power of a natural number? If it is true, is it possible to generalize?
I can use only Euclidean Division. I am not suppose to use the Fundamental theorem of arithmetic.
I think the answer is no, then I have assumed that $m(m+1)=a^{7}$ and I would like to get a contradiction. I know that $m(m+1)=2(1+2+\cdots+m)$, but I got stucked here. I have tried to use the Euclidean Division and write $m=aq+r$, for $0\leq r <a$, and this doesn't help me as well.
I would appreciate your help.
 A: Step 1
At first notice that $m$ and $m+1$ are co-prime.
Step 2 use Fundamental theorem of arithmetic to prove
$$m=k^7$$
and
$$m+1=l^7$$
Then substract
$$1=l^7-k^7>1$$ for nontrivial $l,k$
A: Suppose that $x,y$ are co-prime positive integers, $z$ is a positive integer such that $xy=z^m$, where $m$ is a positive integer.
Let $d=\gcd(x,z)$, we'll prove that $x=d^m$ and $y=(d/z)^m$.
Let $x=x_1d$ and $z=z_1d$, we have $x_1y=z_1^md^{m-1}$. For $d|x$ and $\gcd(x,y)=1$, we have $\gcd(d,y)=1$, thus $\gcd(d^{m-1},y)=1$. Notice that $d^{m-1}|x_1y$, we have $d^{m-1}|x_1$. $\gcd(x_1,z_1)=1$ as $d=\gcd(x,z)$, therefore $\gcd(x_1,z_1^m)=1$. As $x_1|z_1^md^{m-1}$, we have $x_1|d^{m-1}$. Thus we obtain $x_1=d^{m-1}$, and $x=d^m,y=(d/z)^m$.
Two lemmas are used:
Lemma1 Given that $\gcd(x,y)=1$, we have $\gcd(x,y^m)=1$, where $m$ is a nonnegative integer.
Lemma2 (Euclid's lemma) Given that $\gcd(x,y)=1$ and $x|yz$, we have $x|z$.
These two lemmas are proved through Bezout's identity. Could you prove alone?
Then, clark's proof works well.
