# 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.

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Are you trying to show that for no $m$, one can write $m(m+1)$ as a seventh power? Otherwise, what is wrong with letting $m=1$? – Alex Becker Aug 17 '12 at 0:49
$m$ and $m+1$ are coprime, so each of them would have to be a seventh power... – anon Aug 17 '12 at 0:49
Or $2(2+1)$... many counter-examples. – Sigur Aug 17 '12 at 0:50
@AlexBecker: The exercise (in Portuguese) begins with "Let $m$ be a natural number". – spohreis Aug 17 '12 at 0:54
Is it supposed to read: Can one write $m(m+1)$ as a seventh power? That is a question. When you write One can write... it sounds like you are making the assumption, or refer to a known fact that it can be written as a seventh power, which doesn't make any sense! The order of the words changes the interpretation of that sentence. – Jyrki Lahtonen Aug 17 '12 at 8:22

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 $l,k$

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I am not suppose to use the Fundamental theorem of arithmetic. – spohreis Aug 17 '12 at 0:55

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.

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