[QUOTE=science_man_88;457771]sorry typo 4^101-1 equals (4^2)^50*(4^1) -1 which reduces to 1^50*(4^1)-1 = 1*(4^1)-1 which is just 4^1-1 = 3 is divisible by 3.[/QUOTE]

Is used to determine whether the 3 factor is not 5^101-4^101,?I still do not understand.I'm not a math major.

[QUOTE=353085177;457773]Is used to determine whether the 3 factor is not 5^101-4^101,?I still do not understand.I'm not a math major.[/QUOTE]

neither am I I've just been on this forum for closing in on 8 years. well yes you can because you can prove something about each part: [TEX]5^{101} -4^{101} \equiv 2^{(101%2)}-1^{(101%2)} \pmod 3 \equiv 2^1-1^1 \equiv 1 \pmod 3[/TEX]

[QUOTE=science_man_88;457776]neither am I I've just been on this forum for closing in on 8 years. well yes you can because you can prove something about each part: [TEX]5^{101} -4^{101} \equiv 2^{(101%2)}-1^{(101%2)} \pmod 3 \equiv 2^1-1^1 \equiv 1 \pmod 3[/TEX][/QUOTE]

But any factor p of (a+1)^n-a^n must be of the form 2kn+1,n%(p-1)=?

[QUOTE=353085177;457785]But any factor p of (a+1)^n-a^n must be of the form 2kn+1,n%(p-1)=?[/QUOTE]

the remainder of n or divisions by p-1 is what n%(p-1) is. Right, ... But, you can use the same style of argument. if a is greater than n then you can take the remainder by this value p so you could reduce the base for any divisor talked about in these cases.

[QUOTE=science_man_88;457787]the remainder of n or divisions by p-1 is what n%(p-1) is. Right, ... But, you can use the same style of argument. if a is greater than n then you can take the remainder by this value p so you could reduce the base for any divisor talked about in these cases.[/QUOTE]

P is the form 2kn+1, n% (p-1) =0, the final results are = 0 (mod p)……

[QUOTE=353085177;457788]P is any prime number, n% (p-1) =0, the final results are = 0 (mod p)……[/QUOTE]

okay, so that extension doesn't apply if n<p which if p=2*k*n+1 is always true. but you can still reduce the bases if they are larger than p.

[QUOTE=science_man_88;457789]okay, so that extension doesn't apply if n<p which if p=2*k*n+1 is always true. but you can still reduce the bases if they are larger than p.[/QUOTE]

But if p=2*k*n+1,n must be less than p

[QUOTE=353085177;457791]But if p=2*k*n+1,n must be less than p[/QUOTE]

okay but in the case of 505^101-504^101 you can show this has remainder 183 on division by 203 because that's what 99^101-98^101 is so the reducing the base part can still apply even for these numbers. I guess I don't see what's new in your page.

It's time to move to Misc.Math.

[QUOTE=Batalov;457805]It's time to move to Misc.Math.[/QUOTE]

agreed.

the whole thing divides by 2*a+1 when n is odd ( (a+1)^n+a^n , I mean by whole thing), as a+1 can be restated in the remainder math as -a, giving (-a)^n+(a^n) with this modulus ( as it's called). so (5^101+4^101) divides by 9. the difference side of things (a+b)^n-a^n is always divisible by (a+b)-a=b so only if b is 1 is there not a trivial factor.