(2 + 3)(2^2 + 32)(2^4 + 3^4)(2^8 + 3^8)(2^16 + 3^16)(2^32 + 3^32)(2^64 + 3^64) is equivalent to 2^x + 3^y. Compute x and y.
\((2 + 3)(2^2 + 3^2)(2^4 + 3^4)(2^8 + 3^8)(2^{16} + 3^{16})(2^{32} + 3^{32})(2^{64} + 3^{64})\\ \color{blue}=1.17901845777\cdot 10^{61}=2^x+3^y\)
I was looking for x and y, not the actal answer.
I don't know of any solution.
According to Wolfram alpha
https://www.wolframalpha.com/input/?i=%282+%2B+3%29%282%5E2+%2B+3%5E2%29%282%5E4+%2B+3%5E4%29%282%5E8+%2B+3%5E8%29%282%5E16+%2B+3%5E16%29%282%5E32+%2B+3%5E32%29%282%5E64+%2B+3%5E64%29%3D2%5Ex%2B3%5Ex
You can expand it out. Replace the 2 with a and the 3 with b
I suppose you could just expand it out but there is probably a better method
\((a^{2^0}+b^{2^0})(a^{2^1}+b^{2^1})(a^{2^2}+b^{2^2})(a^{2^3}+b^{2^3})(a^{2^4}+b^{2^4})(a^{2^5}+b^{2^5})\\\)
