# Fermat Was Wrong? Puzzler

From Car Talk.

RAY: Here it is. Everyone, almost everyone remembers from his or her days in school the Pythagorean Theorem.

TOM: Yes.

RAY: A squared, plus B squared, equals C squared. And there are numbers like three, four and five; five, 12, 13 which satisfy that little equation. And many hundreds of years ago a French mathematician by the name of Fermat said, this only works for squares. He said, if you take A, B, and C, integers A, B, and C…

TOM: Yes.

RAY: And there are some A squared plus B squared that will equal C squared, and we believe that. We know we have verification of it. We got real numbers that work.

TOM: Right.

RAY: He said, if it isn’t squared but it’s something else like cubes or to the fourth power or to the fifth power, it doesn’t work. So, for example, there is no A cubed plus B cubed, which equals C cubed. There is no A to the fourth plus B to the fourth that equals C to the fourth.

As luck would have it, a young mathematician issues a statement that he has three numbers which prove Fermat’s theorem is incorrect. He calls a press conference. Now, he doesn’t want to divulge everything right away. He wants to dramatize, build a little bit, does he not?

TOM: Gonna give them one number.

RAY: He gives them all three numbers. He doesn’t tell the power.

TOM: Ah!

RAY: He’s going to give them A, B, and C. Here are the numbers, you ready?

A equals 91. B equals 56. C equals 121.

So, it just so happens that at this little impromptu press conference, there are all these science reporters from all the po-dunky little newspapers that are around this town. And one of the guys, one of the reporters has his 10-year-old kid with him, because this happens to be a holiday. He’s off from school. And the kid very sheepishly stands up and raises his hand, and he said, I hate to disagree with you, sir, but you’re wrong.

The question is, how did he know?

Love it.

## 6 thoughts on “Fermat Was Wrong? Puzzler”

1. Argh, I have no idea … “Notify me of follow-up comments via e-mail.”

1. The numbers they gave are important. Another set of numbers that could have been used are 7765836, 6491031, and 8091126. I wouldn’t want to try and raise those numbers to different powers, so try simplifying the problem?

2. I think I figured it out. Whenever you take an number ending in 1 (like 91) and raise it to a power it will still end in 1. Same with 6. So 91^n+56^n will always end in 7. 121^n will always end in 1. therefore they will never be equal.

3. Jeremy says:

I’m not to good at math but, the sum of A and B is greater than C, right? And the problem is that A, B, and C all have to be raised to the same power, i.e. the power of 3 or 4 or 1,567,875, right? So, how could they ever be equal?

4. Another Dan says:

I like the ones digit solution. Here’s what I thought of: 121 is 11^2, so any power will be 11^2n. Meanwhile, 56 is 7*8 and 91 is 7*13, so A^n + B^n will be divisible by 7. Powers of 11 would clearly not be (from prime decomposition).

@jeremy — No. Consider some right triangles: (3,4,5) ; (5,12,13). Maybe what you were thinking of is the case where C is larger than A + B. In that case C^2 > A^2 + B^2, and it would just get greater with higher powers.

5. If I am correct, you should really be thinking about properties of triangles, specifically the side lengths of triangles. I don’t want to spoil it, but remember, a ten year old boy figured it out.