- #1

- 36

- 2

(B^n + C^n) MOD A = 0

A^n MOD (B+ C) = 0

(A^n - B^n) MOD C = 0

C^n MOD (A - B) = 0

(A^n - C^n) MOD B = 0

B^n MOD (A - C) = 0

The only way I can find is when A = B + C and n is any odd integer

- I
- Thread starter Terry Coates
- Start date

- #1

- 36

- 2

(B^n + C^n) MOD A = 0

A^n MOD (B+ C) = 0

(A^n - B^n) MOD C = 0

C^n MOD (A - B) = 0

(A^n - C^n) MOD B = 0

B^n MOD (A - C) = 0

The only way I can find is when A = B + C and n is any odd integer

- #2

- 16,559

- 6,942

Hm ... I can't see any other way.

- #3

- 36

- 2

Perhaps there is a way with much higher powers and/or of A.

Nice if this way can be proved to be the only way.

- #4

- 36

- 2

Since no replies so far to show that there are other ways with n odd (and greater than one), it looks as if there are none, in which case here is a simple proof of FLT with n odd.

- #5

berkeman

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Hmmm....in which case here is a simple proof of FLT with n odd.

Thread closed temporarily for Moderation...

- #6

mfb

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Are ##n^{17} + 9## and ##(n+1)^{17} + 9## always coprime? If you check all n starting from 1 you'll never find a counterexample. Does that prove there is no counterexample? No, and indeed the claim is wrong: n=8424432925592889329288197322308900672459420460792433 is the first counterexample.

We know that there are no counterexamples to Fermat's last theorem of course, as there is a proof. But that doesn't mean you can prove it in different ways just by looking for counterexamples, failing to find them (of course you won't find any) and then just stopping somewhere. You have to independently prove that there cannot be any counterexamples. As this thread does nothing in that aspect, it will stay closed.

- #7

Mark44

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An even simpler example:You cannot prove a general rule by looking for small counterexamples.

Question: Is ##x^2 + x + 41## prime, with x an integer value?

If x = 1, we get ##1^2 + 1 + 41 = 43## -- prime

If x = 2, we get ##2^2 + 2 + 41 = 47## -- prime

If x = 3, we get ##3^2 + 3 + 41 = 53## -- prime

If x = 4, we get ##4^2 + 4 + 41 = 61## -- prime

If x = 5, we get ##5^2 + 5 + 41 = 71## -- prime

If x = 6, we get ##6^2 + 6 + 41 = 83## -- prime

We can continue with this exercise for many more values of x, producing a prime at each step. It would

However, if x = 40, we get ##40^2 + 40 + 41 = 40^2 + 40 + 40 + 1 = 40^2 + 2 \cdot 20 + 1 = (40 + 1)^2##, which is a perfect square, so obviously not prime.

To rephrase what @mfb said, "No specific number of examples will suffice to prove a general rule."

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