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

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?

• You have no answered to my WARNING. Your problem admits an infinity of solutions. Even if you restraint with x and y be integers you have to solve a diophantine equation before say x = y only. With x and y being reals you have all the solution you could desire for x-y.Regards. (sorry for bad English). – Piquito Apr 25 '15 at 12:44

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 $x-y=0$

• Brilliant answer! I never would have thought to multiply by $1$ – user170231 Apr 23 '15 at 2:35
• @user170231 the key is not multiply by $1$ but the recognition that the product is a telescopic one. If a sum/product appear in an exam/contest which looks incredibly complicated to the point impossible to solve in the time allowed. then you should check whether it is a telescopic one. There are not too many tricks to setup complicated looking but easy to solve problem. – achille hui Apr 23 '15 at 2:45
• Okay, I see what you mean. If we were given $\prod (8^{2k}+6^{2k})$, we could still use the same approach, but we would have to divide by $2$, right? – user170231 Apr 23 '15 at 2:53
• @user170231: Yes – freak_warrior Apr 23 '15 at 3:06

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.