IIT JEE , AIEEE Forum - Mathematics , Algebra

shreyaarya

prove fermat's theorem

rudra.panda

are you asking the proof of

if p is a prime number then $(a_1+a_2+a_3+\cdots+a_n)^p\equiv (a_1^p+a_2^p+\cdots+a_n^p)\mod\ p$ ?

Here it is:

$(a_1+a_2+a_3+\cdots+a_n)^p=(a_1+b_1)^p$

where $b_1=a_2+a_3+\cdots+a_n$ $\equiv(a_1^p+b_1^p)\mod p$

$\equiv(a_1^p+b_1^p)\mod p$

$\equiv[a_1^p+(a_2+a_3+\cdots+\a_n)^p]\mod p$

$\equiv[a_1^p+(a_2+c_2)^p]\mod p$ where $c_2=a_3+a_4+\cdots+a_n$

$\equiv (a_1^p+a_2^p+c_2^p)\mod p$

continuing like this we get $(a_1+a_2+a_3+\cdots+a_n)^p\equiv (a_1^p+a_2^p+\cdots+a_n^p)\mod\ p$

Or if you just want the proof of $(a+b)^p\equiv (a^p+b^p)\mod p$