equation and this common factor can be removed. Besides, one and only one
of the numbers, x, y or z must be even as they are pair wise relatively prime while the
sum of two odd numbers must be even. Euler’s Proof of the case n=3
First, let x, y and z be the solutions for x3+y3=z3
where x, y and z are pair wise relatively prime since any factor that divide two of the
solutions would certainly divide the third by virtue of the equation and this common factor can be removed. Besides, one and only one
of the numbers, x, y or z must be even as they are pair wise relatively prime while the
sum of two odd numbers must be even.
Now suppose x and y are odd and z is even. Then x+y and x-y are both even, say 2p and 2q respectively, and x=1/2(2p+2q)=p+q, y=1/2(2p-2q)=p-q. x3+y3 can be factorized as (x+y)(x2-xy+y2)=2p[(p+q)2-(p+q)(p-q)+(p-q)2] =2p(p2+3q2).
Leonhard Euler (1707-1783)
p and q are of opposite parities as p+q and p-q are odd. They are also relatively prime because their common factor would also be the common factor p+q and p-q, which contradicts that fact that x=p+q and y=p-q are relatively prime. Hence the common factor could only be 1.
We can assume that p and q be positive as we can interchanged the position of x and y if x<y to give q>0. Also the case x=y is impossible because then x=y=1, z3=2. As by assumption, x and y are both odd and p and q are positive integers of opposite parity so that 2p(p2+3q2)=cube (there is no harm if z is odd and one of x or y is even because in this case the odd one, say y3, can be moved to the right side
x3=z3-y3=(z-y)(z2+zy+y2)
Using the same reasoning as in the first case where z is even while x and y are odd, we can get x3=2p[(q+p)2-(q+p)(q-p)+(q-p)2] =2p(p2+3q2)
where p, q are relatively prime positive integers of opposite parity.
If 2p and p2+3q2 are relatively prime and for their product to be a cube, we can conclude that each of them must be a cube(there is a theorem which states that when the product of mn where *gcd (m,n)=1 is a power ath to another number say g, so that mn=ga, then each of m and n must equals to a power ath to some other integers, that is m=ca and n=da for some integers c and d.) However, the statement that 2p and p2+3q2 are relatively prime is not quite justified. Since p and q have opposite parity, p2+3q2 is odd and any common factor of 2p, p2+3q2 would be a common factor of p, p2+3q2 and therefore the common factor of p, 3q2. Since p and q are relatively prime, the only possible common factor of p, 3q2 is 3 so that 2p and p2+3q2 are not relatively prime. So now we can split the proof into two case so that in case 1 we assume that 2p, p2+3q2 are relatively prime while case 2 we assume that there is a common factor for 2p and p2+3q2 which is 3.
Case 1
Since we assume that there is no common factor between 2p and p2+3q2, 2p and p2+3q2 are both cubes as concluded above.
Using the formula
(a2+3b2)(c2+3d2)=(ac-3bd)2+3(ad+bc)2
We can find cubes of the form p2+3q2 by writing
(a2+3b2)3=(a2+3b2)[ (a2-3b2)+3(2ab)2]
=[ a(a2-3b2)-3b2(2ab)] 2+3[ a(2ab)+b(a2-3b2)]
=(a3-9ab2)2+3(3a2b-3b3)2
Therefore, we can find cubes of the form p2+3q2 by choosing a, b at random and to se p=a3-9ab2 and q=3a2b-3b3 so that p2+3q2=(a2+32)3.
(Well the gap of Euler’s proof here is to proof that this is the only way for p2+3q2 to be expressed as a cube so that whenever p2+3q2 is a cube then there must exist a, b such that p and q are given by the above equations. However, Euler based this conclusion on the erroneous argument described earlier.) Now if this conclusion is proved to be correct, then we can reason step by step and reach the conclusion that the case of n=3 of Fermat’s Last Theorem is proved.
Let’s continue our proof (more accurately, Euler’s proof)
Factorize the expressions p= a3-9ab2 so that p=a(a-3b)(a+3b) and q=3a2b-3b3 =3b(a-b)(a+b)
a and b are relatively prime because any common factor of them will divide both p and q and contradicts the assumption that they are relatively prime.
2p can now be expressed in terms of a and b so that 2p=2a(a-3b)(a+3b)=cube.
a and b must have opposite parities otherwise p and q would be both even. As a result, a-3b, a+3b are both odd and the only possible common factor of 2a, a ±3b would be common factors of a, a± 3b and therefore of a, ± 3b. Similarly, any common factor of a+3b and a-3b would be a factor of a and of 3b. We can then see that the only possible common factor is 3 but 3 does not divide a because it would then divide p, hence contradicting the assumption. Therefore 2a, a-3b, a+3b are relatively prime and all three of them must be cubes, so that 2a=a 3, a-3b=b 3, a+3b=g 3. Then b 3+g 3=2a=a 3 and this gives a solution of x3+y3=z3 in smaller numbers than the original solution.
As a 3b 3g 3=2a(a-3b)(a+3b)=2p, which is positive and a factor of z3 if z is even and a factor of x3 if x is even. Hence a 3b 3g 3 is less than z in any case. In fact, a , b , g could be negative, but since (-a )3= -a 3 and negative cubes can be moved to the opposite side of the equation to become positive cubes and the resulting equation is of the form X3+Y3=Z3 in which X, Y, and Z are all positive and Z3<z3. Using the same reasoning, we can again get another set of solutions x,y and z that satisfies the equation x3+y3=z3 where x, y, z are positive and again z3<Z3. The process can continue for infinite number of times to get infinitely many smaller and smaller value of solutions for the equations. However, for positive integers, it cannot decrease indefinitely. An infinite descent has been set up. Hence we can conclude that the case1 of the case n=3 of Fermat’s Last Theorem is proved.
Case 2
In case 2, we assume that e is a factor of p. Then p=3k, for an integer s and 3 does not divide q. Then 2p(p2+3q2)=32x2k(3k2+q2). 32 and 3k2+q2 are relatively prime (still remember that 3 is not a factor of q?) hence both of them are cubes. Again, we have to first proof that for 3k2+q2 to be a cube, q must = a(a-3b)(a+3b) and k=3b(a-b)(a+b) for some integers a, b. Since 32*2k is a cube, 33x2b(a-b)(a+b) is a cube and therefore 2b(a-b)(a+b) is a cube. We can see that the factors are relatively prime. Now take 2b=a 3, a-b=b 3, a+b=g 3, a 3=2b=g 3-b 3 and hence a 3+b 3=g 3 which is of the form X3+Y3=Z3 with Z3<z3. Once again, an infinite descent is accomplished and the theorem is proved.
*gcd(a,b)=greatest common factor of a and b.