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This is taken from the Car Talk puzzler of the week, seen here: http://www.cartalk.com/content/mathematic-mistake-0?question

I'll summarize it thusly:

A hotshot mathematician calls a press conference because he's found a counterexample to Fermat's Last Theorem (which claims that $A^x + B^x = C^x$ has no integer solutions for $A, B$ and $C$ when $x > 2$). However, just to be dramatic (and annoying), he doesn't reveal the whole counterexample, but just the values of $A, B$ and $C$, which are 91, 56 and 121, respectively. The 10-year-old child of one of the reporters attending the press conference raises his hand, and says "Sorry, sir, but you're wrong."

The question is: How did the child know?

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    $\begingroup$ Hint: this would also work with 21, 16, and 41. $\endgroup$
    – user641
    Sep 25, 2013 at 16:44
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    $\begingroup$ To answer the question in the title, the mistake is to call a press conference before you've checked your solution on a computer. $\endgroup$
    – user64687
    Sep 25, 2013 at 16:47
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    $\begingroup$ None of 91, 56, and 121 is prime. $\endgroup$ Sep 25, 2013 at 17:27
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    $\begingroup$ The child had just discovered a remarkable proof of the semi-stable case of the Taniyama-Shimura conjecture, and remarked that Fermat's Last Theorem followed as a corollary. Hence the mathematician must have been wrong. $\endgroup$ Sep 25, 2013 at 22:31
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    $\begingroup$ I saw this pop up as a "Hot Question" and thought FLT = Faster than Light Travel. Then I read and learned something. Yay math! $\endgroup$
    – Kyle Kanos
    Sep 26, 2013 at 2:58

3 Answers 3

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7 divides both 91 and 56, but not 121.

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  • $\begingroup$ -1 if I had the reputation. Since we're adding A^x and B^x, it doesn't matter whether A, B and C share factors. It only matters whether (A^x + B^x) and C^x have the same factors for some x. $\endgroup$
    – Kevin
    Sep 25, 2013 at 19:32
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    $\begingroup$ If 7 | x and 7 | y, then 7 | x^n and 7 | x+y. So 7 | x^n + y^n. Finally, if 7 !| z then 7 !| z^n. So here it's perfectly fine to say that since 7 | A, B and 7 !| C, A^n + B^n != C^n for all n. $\endgroup$
    – Hovercouch
    Sep 25, 2013 at 19:42
  • $\begingroup$ @Kevin, Hovercouch's comment above was directed at you, I believe. $\endgroup$
    – LarsH
    Sep 25, 2013 at 20:22
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    $\begingroup$ I stand corrected. $\endgroup$
    – Kevin
    Sep 25, 2013 at 20:35
  • $\begingroup$ This is the simplest answer. $\endgroup$ Sep 26, 2013 at 5:37
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The child could have plugged $121^2 - 56^2 - 91^2$ into a calculator and seen that it's greater than 0, so there's no way in hell it'd work for $x > 2$.

Or he could have realized there was something off about the last digits of the numbers. What's so special about $1$ and $6$?

EDIT: You don't even need any clever tricks for this one, just a good sense of estimation. $121^2 > 120^2 = 14400$. $56^2 < 60^2 = 3600$. $91^2 < 100^2 = 10000$. Since $10000+3600 < 14400$, then $91^2 + 56^2 < 121^2$.

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    $\begingroup$ Two very neat reasons in the same answer can only make for +1. Sorry about that. $\endgroup$
    – Did
    Sep 25, 2013 at 16:50
  • $\begingroup$ Many thanks! I've got the solution, after discarding my original hypothesis about one of the numbers being a prime. $\endgroup$
    – taserian
    Sep 25, 2013 at 16:59
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$(10a+1)^n$ always ends in $1$ and $(10b+6)^n$ always ends in $6$ so $(10a+1)^n+(10b+6)^n$ always ends in $7$.

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    $\begingroup$ Funny, after reading Hovercouch's answer, I had done the calculation modulo 5 instead of modulo 10. $\endgroup$
    – Carsten S
    Sep 26, 2013 at 10:50
  • $\begingroup$ I believe this is the way a 10 year old would think. $\endgroup$
    – N.S.JOHN
    Apr 20, 2016 at 17:26
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    $\begingroup$ A smart ten year old. $\endgroup$ Apr 20, 2016 at 17:40

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