4
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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). $\endgroup$ – Piquito Apr 25 '15 at 12:44
9
$\begingroup$

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 $x-y=0$

$\endgroup$
  • $\begingroup$ Brilliant answer! I never would have thought to multiply by $1$ $\endgroup$ – user170231 Apr 23 '15 at 2:35
  • 2
    $\begingroup$ @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. $\endgroup$ – achille hui Apr 23 '15 at 2:45
  • 2
    $\begingroup$ 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? $\endgroup$ – user170231 Apr 23 '15 at 2:53
  • $\begingroup$ @user170231: Yes $\endgroup$ – freak_warrior Apr 23 '15 at 3:06
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.