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The equation $a^7xy-a^6y-a^5x=a^4(b^4-1)$ is equivalent to the equation $(a^mx-a^n)(a^py-a^2)=a^4b^4$ for some integers $m$, $n$, and $p$. Find $mnp$.

 Dec 16, 2019
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I am so sorry for being this late, but I recently had to solve this problem too. 

 

If you look at the equation \((a^mx-a^n)(a^py-a^2)=a^4b^4\), you may notice that you can factor the previous equation to be in this form. 

First, you distribute \(a^4\)into the parentheses to get \(a^4b^4 - a^4\). If you look at the other equation, you may see that their answer is very similar to \(a^4b^4 - a^4\). So, you first add \(a^4\) to both sides to get \((a^3x-a^2)(a^4y-a^2)=a^4b^4\). Factoring the left side gives \((a^3x-a^2)(a^4y-a^2)=a^4b^4\). So, \((m,n,p)=(3,2,4)\), which means \(mnp=3\cdot2\cdot4=\boxed{24}\).

 Mar 18, 2021

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