Let’s have the equation $a^k+\frac{k \cdot a^k}{x \cdot y^n}=m^k$ where $k≥2$ and $x≠y$ and $a, k, x, y, n, m$ positive integers greater than zero. If $x \cdot y^n = f \cdot a^g$ where $f$ and $g$ any positive integer greater than zero then the equation $ a^k + \frac{k \cdot a^k}{x \cdot y^n}=L$ has infinite solutions where L is an integer but never has the form $L=m^k$. Does anyone know what will be the easiest way to prove that the above equation is impossible for all values $k≥2$?
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It's elementary: let $d=(a,m)$, $a=a_1d$ and $m=m_1d$. Then dividing both sides by $d^k$ gives $a_1^k+\frac{ka_1^k}{xy^n}=m_1^k$. Let $p$ be any prime divisor of $a_1$. Since $(a_1,m_1)=1$, $p\nmid m_1$, so $p\nmid \frac{ka_1^k}{xy^n}$. In particular, the number of $p$ factors in $a_1^k$ is no greater than the number of $p$ factors in $xy^n$. Since this holds for all prime divisors of $a$, we know that $a_1^k\mid xy^n$. In particular $a_1^k\le xy^n$. Then the left hand side $\le a_1^k+k$. On the other hand, clearly $m_1>a_1$, so $m_1\ge a_1+1$, so $m_1^k\ge (a_1+1)^k\ge a_1^k+ka_1+1\ge a_1^k+k+1$ (here we used $k\ge2$, so that the binomial expansion of $(x+1)^k$ contains at least the terms $x^k$, $kx$ and 1.) Now we have a contradiction.
