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# Fermat's Last Theorem (FLT)

Fermat's Last Theorem (FLT), a significant hypothesis in number theory , was first stated by Pierre de Fermat, a 17th-Century laywer and amateur mathematician. The proposition was discovered by his son Samuel while collecting and organizing the elder Fermat's papers and letters posthumously.

The proposition is as follows. Suppose we have the following equation:

x n + y n = z n

where x , y , and z are nonzero integer s. Then the equation has no solution for integers n larger than 2.

Fermat did not state a proof of this hypothesis, although he said that he had found a remarkable demonstration but did not have space in the margin of his text to write it down. Mathematicians immediately began seeking a proof. (Many mathematicians today doubt that Fermat had actually found a valid proof.) The hypothesis was demonstrated true for increasingly large values of n , but proving the theorem in general, for all integers n greater than 2, remained elusive for centuries. Over the next three hundred years, mathematicians from all over the world sought to prove Fermat's Last Theorem; it was considered by many to be the Holy Grail of mathematics.

Two strategies of proof can reasonably be tried. First, one can assume that the equation has a solution for some nonzero integers x , y , and z , and for some n greater than 2, and then derive a contradiction from this assumption. This tactic is formally known as reductio ad absurdum. Second, one might prove that the equation has no solution for n = 3, and then demonstrate that if the equation has no solution for n = k , where k is an unspecified integer, then there exists no solution for n = k + 1. This is the technique of mathematical induction.

In the 1990s, the British mathematician Andrew Wiles produced a proof of FLT that, after some refinements, has withstood all challenges to date.

This was last updated in March 2011

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Fermat's Last theorem is proved by taking the difference of two binomial expansions
(p+q)^3-(p-q)^3, being 6p^2.q+2q^3, and recognising that 6+2=8 which is 2^3, so that the full expression will only have a cube root if p=q, which nullifies the second term of the original difference, leaving the tautology (2q)^3=8.q^3.
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