FERMAT S LAST THEOREM REVISITED AGAIN Colin Newton 8/0/08 0
FERMAT S LAST THEOREM When Fermat wrote a note in the margin of his coy of Bachet s Arithmetica to the effect that he had a marvellous roof that x n + y n z n where n is any integer greater than, erhas he was thinking along the following lines. The argument is based on the roof that if y and z are nth owers then x is not an nth ower, conversely if both x and y are nth owers then z is not an nth ower. The roofs investigate owers of n, n + and n for all n > 0. Definition Define a Pythagorean Trile (PT) as x + y = z, and all three are integers. Definition Define a Diohantine Trile (DT) as y x = zd, and all three are integers. Theorem Proof Any integer x > with ( w < x) x generates all Pythagorean Triles. x + y = z and without loss of generality assume x < y then x = z y = (z y)(z + y) since (z y)(z + y) = z + z y y z y = z y. Let z y = w then z = y + w and x = w (y + w + y) therefore x = w (y + w) = yw + w hence yw = x w and y = x w and z = y + w = x w + w = x + w For x >, w x and for y to be ositive, w < x. Notes for Theorem. If x is even, w odd or even, then y, and hence z, may be fractions with the denominator equal to. e.g. for x =, y = 5/ and z = 7/ with w =, multilying by gives the Trile (8, 5, 7 ). x = 0 roduces (0,, 6) with w =, but also (0, /, 9/) with w = and (0, 5/, 5/) with w = 5. Multily by to obviate the roblem. For all Primitive Pythagorean Triles GCD (x, y) = : For x > there always exists a Pythagorean Trile, for x odd let w =, x even let w =. If x is odd, w is odd since the numerators of both y and z must be even to be divisible by. If x is even, and w is even then the numerators of both y and z are divisible by. 5 If x is odd, w is odd and y is even, if x is even, w is even and y is odd, in both cases z is odd. Theorem Any integer x > with ( wd < x) x generates all Diohantine Triles. Proof y x = zd. For zd to be ositive and without loss of generality assume x < y then x = y zd = (y zd)(y + zd) since (y zd)(y + zd) = y + yzd zdy zd = y zd. Let y zd = wd then y = zd + wd and x = wd (zd + wd + zd) hence x = wd (zd + wd) = zd wd + wd therefore zd wd = x wd and zd = x w d and y = zd + wd = x w d + w d = x + w d Note For Theorem below, the x and y in Theorem are equal to the x and y in Theorem.
Theorem Proof. Any integer x > and w such that ( w < x) x cannot generate both a Pythagorean Trile and a Diohantine Trile simultaneously. Assume that x + y = z is a PT, then to generate a DT requires that y x = zd, hence zd = (x w ) w x = (x w ) w x w w = z y from Theorem and zd = x 6x w + w zd an integer. To be square the exression under the must be of the form x x w + w = (x w ) by the Binomial Theorem, i.e. the coefficients of the exansion of (x w ) are given by the nd row of Pascal s Triangle. Assume that y x = zd is a DT, then to generate a PT requires that x + y = z hence z = (x + w d) w + x = (x + w d) + w dx d w wd = y zd from Theorem d and z = x + 6x w d + w d z an integer. To be square the exression under the must be of the form x + x wd + wd = (x + wd ) by the Binomial Theorem, i.e. the coefficients of the exansion of (x + wd ) are given by the nd row of Pascal s Triangle. Before roceeding, a coule of Lemmas (Lemmata?) and also a coule of identities. Lemma The roduct of two squares is also a square. Proof x y = (x y) for all x and y greater than 0. Lemma Proof Identity Identity Theorem The roduct of two numbers, where one or both are non-square is not a square. x y = (x y) x y square for one or both x and y > and square. (x + y) n = c0x n + cx n- y + cx n- y +... + cx y n- + c0y n by the Binomial Theorem where the coefficients c0, c, c,... c, c, c0 are the values of the corresonding numbers in the nth row of Pascal s Triangle and c0 =. x n y n = (x y) (x n- + x n- y + x n- y +... + x y n- + x y n- + y n- ) where all the coefficients are. It is left to the reader to multily it out to rove it. Fermat s Last Theorem for x + y a fourth ower. Proof The roof is for x or y not being a fourth ower, hence x + y z. Assume for a contradiction, x + y = z then x = z y = (z y ) (z + y ) where both (z y ) and (z + y ) require to be square by Lemma. However, by Theorems, and, if (z y ) = a square then (z + y ) a square and vice-versa, hence (z y ) (z + y ) a square by Lemma, therefore x is not a square and not equal to a fourth ower since x = (x ), this roves x z y hence the equation x + y z is true. Exchanging x and y roduces the same result
Theorem 5 Fermat s Last Theorem for n, i.e. x n + y n z n for all n >. The roof is by induction on n. For n = x + y = z x = z y = (z y ) (z + y ) x by Theorems, and, and already roven by Theorem. For n x n + y n = z n x n = z n y n = z n y n = (z n y n ) (z n + y n ) = (Z Y ) (Z + Y ) where Z = z n and Y = y n but (Z Y ) (Z + Y ) a fourth ower by Theorems,, and. For (n + ) x (n+) + y (n+) = z (n+) x (n+) = z (n+) y (n+) = z (n+) y (n+) = (z (n+) y (n+) ) (z (n+) + y (n+) ) = (Z Y ) (Z + Y ) where Z = z (n+) and Y = y (n+) but (Z Y ) (Z + Y ) a fourth ower by Theorems,, and. Theorems and 5 rove that the equation x n + y n z n is true for all n > 0. Theorem 6 Fermat s Last Theorem for n + where n =, i.e. x 6 + y 6 z 6. then Assume for a contradiction that x 6 + y 6 = z 6 x 6 = y 6 z 6. x 6 = (z y ) (z + z y + y ). To be a sixth ower, the first bracket requires to be a square which it may be by Theorem, but also the second bracket requires to be a fourth ower, which is imossible since to be so, it requires to be of the form (z + z y + y ) by the Binomial Theorem as er Identity. Hence x is not a sixth ower and the equation x 6 + y 6 z 6 is true. Theorem 7 Fermat s Last Theorem for n + where n is any integer >. The roof is by induction on n, the case for n = is roven by Theorem 6 but for n > n + x n+ + y n+ = z n+ x n+ = z n+ y n+ therefore x n + = (z y ) (z n + z n- y +... + x y n- + y n ) where the first bracket may be square by Theorem, however all the coefficients of the second bracket are, but for a fourth ower should be the coefficients of the nth row of Pascal s Triangle by the Binomial Theorem as er Identity. Hence x is not a (n + )th ower and the equation x n+ + y n+ z n+ is true for all n >. (n + ) + x (n+)+ + y (n+)+ = z (n+)+ x (n+)+ = z (n+)+ y (n+)+ therefore x (n+)+ = (z y ) (z (n+) + z (n+)- y +... + x y (n+)- + y (n+) ) where the first bracket may be square by Theorem, however all the coefficients of the second bracket are, but for a fourth ower, require to be the coefficients of the (n + )th row of Pascal s Triangle by the Binomial Theorem as er Identity. Hence x is not a ((n + ) + )th ower and the equation x (n+)+ z (n+)+ y (n+)+ x (n+)+ + y (n+)+ z (n+)+ is true for all n >. Theorems to 7 rove Fermat s Last Theorem is true for all even owers of n greater than.
Theorem 8 Fermat s Last Theorem for n where n is any integer > 0. The roof is by induction on n. For n = n = x + y = z x = z y = (z y) (z + z y + y ) by Identity. The only two factors that satisfy this equation are x and x but (z + z y + y ) square since, by the Binomial Theorem it must be of the form (z + z y + y ) = (z + y). Hence, by Lemma, x z y i.e. x + y z. Note (z y) (z + z y + y ) = z + z y + z y z y z y y = z y. For n x n- + y n- = z n- x n- = z n- y n- and by Identity = (z y) (z n- + z n- y +... + z y n- + y n- ) where the second bracket may be the exansion of several factors. The only two factors that satisfy this equation are x and x n-. The second bracket cannot be a (n )th ower, since it is of the form of Identity and not of the form of Identity, hence x n- + y n- z n- and therefore equation x n- + y n- z n- is true. (n + ) x (n+)- + y (n+)- = z (n+)- x (n+)- = z (n+)- y (n+)- and by Identity = (z y) (z (n+)- + z (n+)- y +... + zy (n+)- + y (n+)- ) where the second bracket may be the exansion of several factors. The only two factors that satisfy this equation are x and x (n+)-. The second bracket cannot be a ((n+)-)th ower, since by the Binomial Theorem, it must be of the form of Identity. Therefore x is not a ((n + ) )th ower and the equation x (n+)- y (n+)- + z (n +)- x (n+)- + y (n+)- z (n+)- is true. Theorem 9 Fermat s Last Theorem for n + where n is any integer > 0. The roof is by induction on n. For n = n + = 5 x 5 + y 5 = z 5 x 5 = z 5 y 5 = (z y)(z + z y + z y + zy + y ). The only two factors that satisfy this equation are x and x but (z + z y + z y + zy + y ) a fourth ower since to be so requires that (z + z y + 6 z y + z y + y ) = (z + y) by the Binomial Theorem. Hence x 5 z 5 y 5 and therefore equation x 5 + y 5 z 5 is true. For n + x n + + y n + = z x n + = z y and by Identity = (z y) (z n + z n - y +... + z y n - + y n ) where the second bracket may be the exansion of several factors. The only two factors that satisfy this equation are x and x n. The second bracket cannot be a nth ower, since it is of the form of Identity and not of the form of Identity, and therefore x n + y n + z and hence equation x n + + y n + z is true. (n + ) + x (n+)+ + y (n+)+ = z (n+)+ x (n+)+ = z (n+)+ y (n+)+ = (z y) (z (n+) + z (n+)- y + z (n+)- y... + z y (n+)- + y (n+ ) where the second bracket may be the exansion of several factors. The only two factors that satisfy this equation are x and x (n+). The second bracket cannot be a (n + )th ower, since it is of the form of Identity and not of the form of Identity, hence x (n+)+ z (n+)+ y (n+)+ which imlies that the equation x (n+)+ + y (n+)+ z (n+)+ is true. Theorems 8 and 9 rove Fermat s Last Theorem is true for all odd owers of n greater than.