31 Jan 08
Originally posted by FabianFnasI experimented with it a bit and it seems likely that you got yourself a theorem. With all numbers I tried for n the result was always the form of a number ending in 0 and numbers that end in 0 are divisible by 5. But I can't get way to write your expresion in a product that has 10 in it.
I found out that 1^5-1 is dividable with 5. So is 2^5-2 and 3^5-3.
Is 2008^5-2008 also dividable with 5?
Is perhaps all n^5-n dividable with 5? Show me this is the case, or not.
This is all I got so far
n^5-n=n*(n^4-1)=n*(n^2+1)*(n^2-1)=n*(n^2+1)*(n+1)*(n-1)
Now n*(n+1) is the result of an arithmetic summation and I can't see if that help us or not but nonetheless it is interesting. And (n^2+1)*(n-1) maybe is something to though I don't know what...
I'll try for a bit longer and tell you about updates.
31 Jan 08
Originally posted by FabianFnasIs 2008^5-2008 also dividable with 5? YES
I found out that 1^5-1 is dividable with 5. So is 2^5-2 and 3^5-3.
Is 2008^5-2008 also dividable with 5?
Is perhaps all n^5-n dividable with 5? Show me this is the case, or not.
2008^5 ends with the digit 8.
(since considering last digits 8*8 = 64, 4*8 = 32, 2*8 = 16, 6*8=48)
Subtracting 2008 gives us a number ending in zero and hence divisible by 5.
There are only 10 cases to consider; numbers ending in 0,1,2, ..9. Prove for each of these and your theorem is proven.
31 Jan 08
1 edit
Proof by induction that for all n >= 0, 5|(n^5 - n)
Base case
let n = 0
0^5 - 0 = 0 is divisible by 5 so result holds
Assume by inductive hypothesis result holds for n = k, ie; 5|(k^5-k)
Let n = k+1
(K+1)^5 - (k+1)
=k^5 +5k^4 +10k^3 + 10k^2 + 5k +1 - (k+1)
=k^5 +5(k^4 +2k^3 +2k^2 + k) - k
=(k^5 - k) + 5(k^4 +2k^3 +2k^2 + k) is divisible by 5 (inductive hypothesis)
Hence result holds for n = k+1
Therefore, by induction on n it is true that for all n >= 0, 5|(n^5 - n)
a similar argument can be used for n < 0
31 Jan 08
1 edit
Originally posted by Tricky DickyIf a not divisable by 5, hmm, what about when a is divisable by 5? Like 12345^5-12345 ?
It's clearly true if the number you start with is divisible by 5.
Fermat says a^4 = 1 (mod 5) if a not divisible by 5.
So a^5 = a (mod 5), whence the result.
But even if you don't know Fermat and what he says you can come to the correct result quite easy.
Edit: ... and now I see that Agerg has came up with a correct result. Well done!
31 Jan 08
1 edit
Originally posted by FabianFnaswell, clearly when a is divisible by 5, a=0 (mod 5) and so a^5=0 (mod 5)... so a^5-a = 0 (mod 5). The nonzero modulo classes are interesting because we can invoke Fermat's Little Theorem, a^p-1 = 1 (mod p) for any prime p. I think this is what Tricky Dicky was saying.
If a not divisable by 5, hmm, what about when a is divisable by 5? Like 12345^5-12345 ?
But even if you don't know Fermat and what he says you can come to the correct result quite easy.
Edit: ... and now I see that Agerg has came up with a correct result. Well done!
In fact, this proves that it is true for ALL prime number exponents, not just 5. Since a^(p-1)=1 (mod p), we see that a^p = a (mod p) and a^p - a = 0 (mod p).
So 12345^71-12345 is divisible by 71... and so forth. Fermat's "Little Theorem" is quite powerful 🙂
Edit: note, similarly to what Tricky Dicky said, a^p-1 = 1 (mod p) is only true when p does not divide a. but if it does, a^p = 0 (mod p) and a = 0 (mod p)... so the result a^p - a = 0 (mod p) still holds
31 Jan 08
I'm so not a mathematician but I think that this problem is simple and obvious. the number raised to any power with the number subtracted is obviously still a multiple of the number. no fancy formulas or anything else is needed, nor is it a theorem, at least any more than close the window there's a draft in here is a theorem.
31 Jan 08
Originally posted by coquetten^p-n is still a multiple of n like you say but it is not obvious at all, at least not to me that n^p-n is a multiple of p.
I'm so not a mathematician but I think that this problem is simple and obvious. the number raised to any power with the number subtracted is obviously still a multiple of the number. no fancy formulas or anything else is needed, nor is it a theorem, at least any more than close the window there's a draft in here is a theorem.
And sometimes on math the most difficult things to prove are the obvious ones. If you know calculus try to prove the intermediat value for yourself. Pretty self-evident and prety obvious. At first sight it even looks like not being a theorem at all. But just try to prove it for yourself.
31 Jan 08
Originally posted by coquetteyou're correct that this is obvious... what ISN'T obvious is that it's a multiple of the POWER you raised it to. that's the interesting fact: 3^71 - 3 is a multiple of 71 (in addition to the trivial fact it is a multiple of 3)
I'm so not a mathematician but I think that this problem is simple and obvious. the number raised to any power with the number subtracted is obviously still a multiple of the number. no fancy formulas or anything else is needed, nor is it a theorem, at least any more than close the window there's a draft in here is a theorem.
31 Jan 08
Originally posted by AetheraelWhat does that ^ symbol mean?
you're correct that this is obvious... what ISN'T obvious is that it's a multiple of the POWER you raised it to. that's the interesting fact: 3^71 - 3 is a multiple of 71 (in addition to the trivial fact it is a multiple of 3)
01 Feb 08
3 edits
Originally posted by coquettebut we are looking for divisibility by the index...not the number raised to the index
I'm so not a mathematician but I think that this problem is simple and obvious. the number raised to any power with the number subtracted is obviously still a multiple of the number. no fancy formulas or anything else is needed, nor is it a theorem, at least any more than close the window there's a draft in here is a theorem.
for example, consider 2^4-2...this is not divisible by 4 yet 2^5 -2 *is* divisible by 5...Yes m^n-m is certainly divisible by m but that is of no interest here....is it divisible by n??!
That is what we are interested in!