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 $mnp$.

Guest Sep 21, 2020

#1**+2 **

**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 \(mnp\).**

\(\begin{array}{|rcll|} \hline \mathbf{a^7xy-a^6y-a^5x} &=& \mathbf{a^4(b^4-1)} \\ a^7xy-a^6y-a^5x &=& a^4b^4-a^4 \\ a^7xy-a^6y-a^5x + a^4 &=& a^4b^4 \\ a^7xy-a^5x -a^6y + a^4 &=& a^4b^4 \quad | \quad \mathbf{a^4b^4 = (a^mx-a^n)(a^py-a^2)} \\ a^7xy-a^5x -a^6y + a^4 &=& (a^mx-a^n)(a^py-a^2) \\ a^7xy-a^5x -a^6y + a^4 &=& a^ma^pxy- a^ma^2x-a^na^py+a^na^2 \\ a^7xy-a^5x -a^6y + a^4 &=& a^{m+p}xy - a^{m+2}x-a^{n+p}y+a^{n+2} \\\\ \text{compare} && \text{compare} \\\\ \mathbf{m+2} &=& \mathbf{5} \\ m &=& 5-2 \\ \mathbf{m} &=& \mathbf{3} \\\\ \mathbf{m+p} &=& \mathbf{7} \\ p &=& 7-m \\ p &=& 7-3 \\ \mathbf{p} &=& \mathbf{4} \\\\ \mathbf{n+2} &=& \mathbf{4} \\ n &=& 4-2 \\ \mathbf{n} &=& \mathbf{2} \\\\ \mathbf{n+p} &=& \mathbf{6} \\ p &=& 6-n \\ p &=& 6-2 \\ \mathbf{p} &=& \mathbf{4} \\\\ mnp &=& 3*2*4 \\ \mathbf{mnp} &=& \mathbf{24} \\ \hline \end{array}\)

heureka Sep 22, 2020