**This article was drafted on 19-09-2011, but not publicized earlier.**

Recently I had given a lot of proofs and corollaries of a new theorem that I called 50cent theorem, in this article:

I will use the general form to prove Fermat’s theorem in just a few line, a feat never acheived.

The idea of the theorem (50cent) is for every integer to it’s k-th power, there is atleast one integer (and unique when the power is close to the prime number in the theorem) in the (k-1)th vicinity of the k-th power of the integer which is an integer multiple of 5(in more generalization of a prime number close to the power k). (k-1)th vicinity means a difference from an integer within the scope of integers -(k-1), -k, .. 0, .., k, (k-1). ie. (a)^k = 5*m + P where P is any integer in the scope -(k-1), -k, .. 0, .., k, (k-1).

SO from the equation above it is clear that any k-power integer of an integer is a 5th multiple plus another integer, always.

Fermat’s Theorem: a^n + b^n = c^n is not valid, for higher powers than n=2.

LHS = 5*m_a + P_a + 5*m_b + P_b = 5*(m_a + m_b) + P_a + P_b, since P_a and P_b are integers in the same scope of integer n-1, they are equivalent although different.

RHS = c^n + P_b,

=> LHS is not equal to RHS unless one of the integer P_a or P_b is zero always.

This is valid till the prime number 5 is sufficient to test the powerfreeness of another integer. {m,n,k} are numbers/integers in the vicinity and for very high powers in Fermat’s theorem one needs higher prime numbers. This theorem has been evidently tested for small powers.

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