Is there another way that these statements can all be true?

  • #1
When n is odd
(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
 

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  • #2
phinds
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Hm ... I can't see any other way.
 
  • #3
I've written a Qbasic program to look for any other way with powers up to 39 and A up to 600 but no other way found. So I conclude that there is no other way.

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
Note if you omit (B^n + C^n) MOD A = 0 and put n = 2 there are all the Pythaqorean triples that satisfy all the other five statements.

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

Thread closed temporarily for Moderation...
 
  • #6
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You cannot prove a general rule by looking for small counterexamples.

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.
 
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  • #7
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You cannot prove a general rule by looking for small counterexamples.
An even simpler example:
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 seem reasonable to conclude from these examples that ##x^2 + x + 41## must be prime.

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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