If $\prod\limits_{k=0}^5(5^{2^k}+6^{2^k})=6^x-5^y$, what is the value of $x-y$?

Question

I think this might be a contest math question, so I'm tagging it as such.

I don't know how to do something like this by hand (or if it's even possible, though I would presume it is if it's from a contest exam). I wrote a script in Mathematica to determine the answer. Here's the code:

t = Table[6^x - 5^y, {x, 1, 200}, {y, 1, 200}];
For[m = 1, m <= 200, m++,
  For[n = 1, n <= 200, n++,
    If[t[[m, n]] == %1, Print[{m, n}]];
     ];
   ];

(where $\%1$ denotes the computed product, approximately $6.33\times10^{49}$) which returns $\{64,64\}$, and so $x-y=0$. How would I do this without the aid of software?

Answer 1

Use the fact that $a^2-b^2 = (a+b)(a-b)$

Multiplying $(6-5)$ on your LHS, we obtain: $$\prod\limits_{k=0}^5(5^{2^k}+6^{2^k})=(6-5)\prod\limits_{k=0}^5(5^{2^k}+6^{2^k})$$ $$=(6^2-5^2)\prod\limits_{k=1}^5(5^{2^k}+6^{2^k})=(6^4-5^4)\prod\limits_{k=2}^5(5^{2^k}+6^{2^k})=...$$

Iterating for all the terms in the products, you should get $x=y=2^6=64$, so $x-y=0$

Answer 2

Each factor of the product with exponent $2^k$ is equal to the difference with exponent 2^(k+1) divided by the difference with the same exponent of the considered factor (because an scholar identity with squares). Then the product finally gives the equation 6^(64) - 5^64 = $6^x$ - $5^y$.

WARNING: there is an obvious solution x = y but it is not unique! Let an arbitrary N = $6^x$ - $5^y$ and take “because we want to" y = 2 then we have a unique solution x of N + 25 = $6^x$ which is not x = 2. There are actually infinitely many solutions for arbitrary N positive, say.