# Mistaken counterexample to FLT; where's the mistake?

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?

• Hint: this would also work with 21, 16, and 41. – user641 Sep 25 '13 at 16:44
• To answer the question in the title, the mistake is to call a press conference before you've checked your solution on a computer. – user64687 Sep 25 '13 at 16:47
• None of 91, 56, and 121 is prime. – Andreas Blass Sep 25 '13 at 17:27
• 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. – John Gowers Sep 25 '13 at 22:31
• I saw this pop up as a "Hot Question" and thought FLT = Faster than Light Travel. Then I read and learned something. Yay math! – Kyle Kanos Sep 26 '13 at 2:58

$(10a+1)^n$ always ends in $1$ and $(10b+6)^n$ always ends in $6$ so $(10a+1)^n+(10b+6)^n$ always ends in $7$.

• Funny, after reading Hovercouch's answer, I had done the calculation modulo 5 instead of modulo 10. – Carsten S Sep 26 '13 at 10:50
• I believe this is the way a 10 year old would think. – N.S.JOHN Apr 20 '16 at 17:26
• A smart ten year old. – marty cohen Apr 20 '16 at 17:40

7 divides both 91 and 56, but not 121.

• -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. – Kevin Sep 25 '13 at 19:32
• 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. – Hovercouch Sep 25 '13 at 19:42
• @Kevin, Hovercouch's comment above was directed at you, I believe. – LarsH Sep 25 '13 at 20:22
• I stand corrected. – Kevin Sep 25 '13 at 20:35
• This is the simplest answer. – Bennett McElwee Sep 26 '13 at 5:37

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 $91^2 + 56^2 < 121^2$.

• Two very neat reasons in the same answer can only make for +1. Sorry about that. – Did Sep 25 '13 at 16:50
• Many thanks! I've got the solution, after discarding my original hypothesis about one of the numbers being a prime. – taserian Sep 25 '13 at 16:59