It says that there is no value of a,b and c, with n>2 and all integer numbers that satisfies this: a^n=b^n+c^n I'm only going to use the cosine theorem. Let's consider three points A, B and C. They form the three sides of a triangle: a, b and c. The sides forms three angles, which can go from 0 to 180 degrees. If one angle, say α , is 180, then the other two are 0, but that doesn't affect the results. Then: a^2=b^2+c^2-2b*c*cosα => a^2=b^2+c^2-2*b*c*cos180 => a^2=b^2+c^2+2*b*c => a^2=(b+c)^2 a=b+c or a=b-c These are two basic principles of geometry, if one point is alligned with two others, their distance is the sum or difference between that point with each of the point. We have just proven the case n=1 for Fermat's last theorem. With the angle equal to 0 we have the same result. Let's consider now that the triangle has one right angle. Then: a^2=b^2+c^2-2*b*c*cos90 => a^2=b^2+c^2 I have just proven the Pythagorean theorem , and the case n=2 for Fermat's last theorem. Cosα is only and integer number if the angles are 0, 90, 180... , and we have just seen that, if the angle is 180, n=1; and if the angle is 90, n=2. For any other values of n, the angle will be between 90 and 0, so cosα will not be an integer number. Fermat's last theorem says that there is no value for a, b and c, with n>2 and all of them being integer numbers that makes this possible: a^n=b^n+c^n If we consider a, b and c the sides of a triangle, then the cosine theorem must apply. If the cosine isn't an integer number, then you aren't going to end up with a,b and c integer numbers. I hope this is well explained and that i have not made mistakes(and sorry if i have not written something right, because i'm spanish). Please say if this could be correct.