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What is Fermat's Last Theorem?

+2 votes
What is Fermat's Last Theorem??
asked Jul 10, 2014 in Math by luongamanda Evoker (610 points)  

2 Answers

+1 vote
Fermat's Last Theorem states that no three integers can complete the equation that \(a^n + b^n = c^n\), when n = greater than two.
answered Jul 10, 2014 by emily-gan Evoker (770 points)  
edited Jul 15, 2014 by l-johnson
good except the clause that reads "where n = greater than zero" should read
"where n is greater than two. "

More to the point, we know that it works for n=1.  E.g.,  \(1^1 + 2^1 = 3^1\).  

We also know the formula \(a^n + b^n = c^n\) works in the case of n=2 for certain integers; otherwise, Pythagorean's Theorem would not exist.  E.g.,  \(3^2 + 4^2 = 5^2\).
+1 vote

Many of us are familiar with the Pythagorean Theorem. This theorem has many proofs.

The Pythagorean Triples (a,b,c) represent the lengths of the two legs and hypotenuse of right triangles and satisfy the Pythagorean Theorem equation

\(a^2 + b^2 = c^2\)

We can find infinitely many triples.  For example, multiply the triplet (3,4,5) by any positive integer and it also satisfies Pythagorean's Theorem

Fermat’s Last Theorem states that there are no positive integers that satisfy the equation with higher degree than 2.  That is, for every triple (a,b,c), the following inequalities are always true.

\(a^3 + b^3 ≠ c^3\)

\(a^4 + b^4 ≠ c^4\)

\(a^5 + b^5 ≠ c^5\)

. . .

\(a^n + b^n ≠ c^n\)

In general, it says that there are no positive integers (a,b,c) that satisfy the equation \(a^n + b^n = c^n\), for any integer n greater than 2.

This math problem was proposed by Pierre de Fermat in 1637, and after more than 300 years and many incorrect proofs published, Andrew Wiles finally solved the enigma in 1995. It took him 8 years to solve the problem!

answered Jul 10, 2014 by l-johnson Evoker (540 points)  
edited Jul 10, 2014 by l-johnson
How did you get those symbols?
Good question.

If you want to show \(a^2 + b^2 = c^2\), enter "\ ( a^2 + b^2 = c^2 \ )" but remove the space between the backslash and parenthesis in both instances.
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