As we'll see it is clear that the same method is used for the solution in the case n = 2k + 1... more As we'll see it is clear that the same method is used for the solution in the case n = 2k + 1 & n = 2k, that is, we use the same logic of the Pythagorean Triad. The proof is based on the Pythagorean triples, i.e. with n = 2 of the FERMAT equation and is the result of the absolute correspondence with the trigonometric equation of the rectangular triangle, proof that is correct and fully compatible.
Every even integer > 2 is the sum of two prime numbers & equivalent Each odd integer > 5 is... more Every even integer > 2 is the sum of two prime numbers & equivalent Each odd integer > 5 is the sum of three prime numbers USING THE SIEVE OF ERATOSTHENES.
It is proved that when n> 2 for c^n + b^n = a^n if n = 2k + 1 or n=2k+2 , k> = 1 at Z +, on... more It is proved that when n> 2 for c^n + b^n = a^n if n = 2k + 1 or n=2k+2 , k> = 1 at Z +, only 2 things can to happen 1st. sinx = 0 and cosx = 1, or 2nd sinx = 1 and cosx = 0..Respectively.If n = 2k, 1st..cosx = + / - 1 and sinx = 0 or 2nd. sinx = + / - 1 and сosx = 0 will apply, Where k> 1 in Z +.Otherwise (sinx) ^ n + (cosx) ^ n <1 or 1=3 0< (sinx) ^ n + (cosx) ^ n <1 .This is simple because if max x =π/4 then for each power of n> = 3 it always results (sinx) ^ n + (cosx) ^ n <1.Οnly in the case n = 2 is equal to 1 as maximum i.e the (sinx) ^ n + (cosx) ^ n =1. That concludes the proof.Andrew Wiles came to the same conclusion with 200 about pages of theory.
As we'll see it is clear that the same method is used for the solution in the case n = 2k + 1... more As we'll see it is clear that the same method is used for the solution in the case n = 2k + 1 & n = 2k, that is, we use the same logic of the Pythagorean Triad. The proof is based on the Pythagorean triples, i.e. with n = 2 of the FERMAT equation and is the result of the absolute correspondence with the trigonometric equation of the rectangular triangle, proof that is correct and fully compatible.
Every even integer > 2 is the sum of two prime numbers & equivalent Each odd integer > 5 is... more Every even integer > 2 is the sum of two prime numbers & equivalent Each odd integer > 5 is the sum of three prime numbers USING THE SIEVE OF ERATOSTHENES.
It is proved that when n> 2 for c^n + b^n = a^n if n = 2k + 1 or n=2k+2 , k> = 1 at Z +, on... more It is proved that when n> 2 for c^n + b^n = a^n if n = 2k + 1 or n=2k+2 , k> = 1 at Z +, only 2 things can to happen 1st. sinx = 0 and cosx = 1, or 2nd sinx = 1 and cosx = 0..Respectively.If n = 2k, 1st..cosx = + / - 1 and sinx = 0 or 2nd. sinx = + / - 1 and сosx = 0 will apply, Where k> 1 in Z +.Otherwise (sinx) ^ n + (cosx) ^ n <1 or 1=3 0< (sinx) ^ n + (cosx) ^ n <1 .This is simple because if max x =π/4 then for each power of n> = 3 it always results (sinx) ^ n + (cosx) ^ n <1.Οnly in the case n = 2 is equal to 1 as maximum i.e the (sinx) ^ n + (cosx) ^ n =1. That concludes the proof.Andrew Wiles came to the same conclusion with 200 about pages of theory.
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