Question

\[\left(5+6\right)\left(5^2+6^2\right)\left(5^4+6^4\right)\left(5^8+6^8\right)\left(5^{16}+6^{16}\right)\left(5^{32}+6^{32}\right)=\left(6^x-5^y\right)\]
\[x-y=...\]

Answer 1

Multiply the LHS by \(1=(6-5)\), then you have
\[\begin{align*}(6-5)\prod_{k=0}^5\left(5^{2k}+6^{2k}\right)&=(6-5)(6+5)\prod_{k=1}^5\left(5^{2k}+6^{2k}\right)\\\\&=(6^2-5^2)\prod_{k=1}^5\left(5^{2k}+6^{2k}\right)\end{align*}\]
From here you can factor out the next term in the product:
\[\begin{align*}(6^2-5^2)\prod_{k=1}^5\left(5^{2k}+6^{2k}\right)&=(6^2-5^2)(6^2+5^2)\prod_{k=2}^5\left(5^{2k}+6^{2k}\right)\\\\&=(6^4-5^4)\prod_{k=2}^5\left(5^{2k}+6^{2k}\right)\end{align*}\]
and so on.

Answer 2

thank you...
I got x = y = 64
so, x-y =0

Answer 3

Yes, I checked with Mathematica and got the same result.