# foil nightmare within mods

• Sep 26th 2007, 10:46 AM
fruitbodywash
foil nightmare within mods
Alright, the problem goes as follows

in mod 5, compute (a+b)^5
in mod 3 compute (a+b)^3
in mod 2 compute (a+b)^2
in mod 4 compute (a+b)^4
in mod 7 compute (a+b)^7

based on results of your findings, what do you thing (a+b)^n is equal to in mod n? Explain in detail.

Now... the a's and the b's represent classes with in the mods, and I've solved all of them out to my satisfaction except mod 4.

i have a^5 + b^5 in mod 5
a^3 + b^3 in mod 3
a^2 + b^2 in mod 2
a^7 + b^7 in mod 7

however, in mod 4, i have a^4 + 6(a^2)(b^2) + a^4
I've tried solving it out several ways, and I can't make mod 4 as pretty as the other mods. Why is this....? Am I doing the foil wrong? because I've tried it several ways to see if I was missing something, and almost thought i had it until i noticed I made a mistake and then got the result that is making it soooo not nice and neat as the others. Please help (Doh)
• Sep 26th 2007, 10:53 AM
topsquark
Quote:

Originally Posted by fruitbodywash
Alright, the problem goes as follows

in mod 5, compute (a+b)^5
in mod 3 compute (a+b)^3
in mod 2 compute (a+b)^2
in mod 4 compute (a+b)^4
in mod 7 compute (a+b)^7

based on results of your findings, what do you thing (a+b)^n is equal to in mod n? Explain in detail.

Now... the a's and the b's represent classes with in the mods, and I've solved all of them out to my satisfaction except mod 4.

i have a^5 + b^5 in mod 5
a^3 + b^3 in mod 3
a^2 + b^2 in mod 2
a^7 + b^7 in mod 7

however, in mod 4, i have a^4 + 6(a^2)(b^2) + a^4
I've tried solving it out several ways, and I can't make mod 4 as pretty as the other mods. Why is this....? Am I doing the foil wrong? because I've tried it several ways to see if I was missing something, and almost thought i had it until i noticed I made a mistake and then got the result that is making it soooo not nice and neat as the others. Please help (Doh)

$\displaystyle (a + b)^6 = a^6 + 3a^4b^2 + 2a^3b^3 + 3a^2b^4 + b^6$

I think the idea to concentrate on here is that the coefficient of each term is
$\displaystyle \frac{n!}{r!(n - r)!}$

Can you think of when this is 0 or not 0 a modulo n?

-Dan
• Sep 26th 2007, 11:03 AM
ThePerfectHacker
It is easy.

Let $\displaystyle \mathbb{F}_q$ be a finite field with charachteritic $\displaystyle p$.
Then,
$\displaystyle (a+b)^{p^n} = a^{p^n}+b^{p^n}\ \forall a,b\in \mathbb{F}_q$.

So,
$\displaystyle (a+b)^p \equiv a^p + b^p (\bmod p)$ for prime numbers.
• Sep 26th 2007, 11:04 AM
fruitbodywash
point well taken, thanks! *chuckles* I guess i was hoping to have it nice and neat, but, I've learned long ago that its not as pretty as one would hope. But, just to make sure, if n is prime, it will be pretty right? but if its not prime.... then, each thing must be foiled out for itself, yes?

*sighs* I just wonder how my teacher wants me to put that into words.

Thanks:)