Fermat's Last Theorem for Cubes Before considering the integer equation x3 + y3 = z3, it's worthwhile to briefly review the simple Pythagorean equation x2 + y2 = z2. For primitive solutions we can assume x,y,z are pairwise co-prime, x is odd and y is even. The usual approach is to re-write the equation as Then, since the two integer factors on the right are co-prime (and since we have unique factorization for integers), they must each individually be squares, so we have z+x = 2u2 and z-x = 2v2 for co-prime integers u,v, (one odd and one even) from which it follows that z = u2 + v2, x = u2 - v2, and y = 2uv. However, there is another approach to solving the Pythagorean equation that makes use of some deeper properties of integers known to Fermat, and that can be generalized to the case of cubes. This alternative approach relies on the fact that numbers of the form X2 + Y2 with gcd(X,Y)=1 can be "factored" uniquely into a product of primes of the same form, and that the representations of composites of this form are generated by applying the identity It's been speculated that Diophantus knew this identity, although he didn't give it explicitly in any of the (surviving) books of "Arithmetica". The first known explicit description was by Abu Jafar al-Khazin (circa 950 AD), and it also appears in Fibonacci's "Liber Quadratorum" (1225 AD). One could argue that this was really the first discovery of complex numbers, in the abstract sense of Hamilton's ordered pairs, because in C the product of (a,b) and (c,d) is (ac-bd,ad+bc). In any case, Fermat knew that only primes of the form 4k+1 are expressible as a sum of two co-prime squares, and those are expressible in only one way. This, combined with the fact that representations of composites are given by the above formula applied to the representations of their factors, enables us to say that if x2 + y2 is a square then the components x,y are given by squaring a number of the form (u2 + v2) using the above identity. As a result we have which of course agrees with our previous solution. Thus, given the theorems about sums of two squares and their unique factorizations that were known to Fermat, this is (arguably) an even more direct solution than the original one, which is perhaps not surprising, since it is essentially employing the field of Gaussian integers, in disguised form. Now let's consider the analogous equation for cubes, i.e., we seek all non-trivial integer solutions of x3 + y3 = z3. Again we consider only primitive solutions, so without loss of generality we can assume x,y,z are co-prime, one even and two odd. Changing signs if necessary we can make x and y odd and z even. Now we define x=u+v and y=u-v where (u,v)=1, and u,v have opposite parity. Substituting into x3 + y3 = z3 gives Since z is even, u must be even and v must be odd. Now we'll consider two cases. First, assume z is not divisible by 3. In this case 2u is co-prime to u2 + 3v2, so both of those factors must be cubes. Thus we have integers co-prime m,n such that In the case of the Pythagorean equation we had a sum of two squares equal to a square, whereas in this case we have a slightly different quadratic form, X2 + 3Y2, equal to a cube. Notice that we can't simply subtract u2 from both sides of (3) and then factor the right hand side, because it is inhomogeneous, i.e., we would have a cube minus a square, which doesn't factor algebraically over the integers. We can, however, proceed to use the second approach, based on factoring the left hand side of (3) into divisors of the same form, provided we know enough about numbers of the form X2 + 3Y2. Happily, it turns out that we have a direct analog for the "Fibonacci identity". In fact, for any integer N we have so we can always multiply together two numbers of the quadratic form X2 + NY2 to give another number of the same form. With k=3 this identity is With this identity in mind, we state and prove several facts about numbers of the quadratic form X2 + 3Y2 which are useful for continuing our search for solutions of x3 + y3 = z3. LEMMA 1: Every prime p of the form 3k+1 divides some integer of the form a2 + 3b2 with (a,b)=1. PROOF: Since u2 + uv + v2 is an equivalent form under the substitution u=b+a and v=b-a, we need only prove that p divides such an integer, with (u,v)=1. Consider where 3k = p-1. Setting v=1 ensures (u,v)=1 and enables us to write The left hand side is divisible by p according to Fermat's Little Theorem for any integer u co-prime to p. Therefore, the right side is also divisible by p for every such u. In order for p to NOT divide any of the number u2k + uk + 1, it must divide each of the numbers uk - 1 for u = 1,2,3,..,p-1. However, the congruence u(p-1)/3 = 1 (mod p) can have no more than (p-1)/3 distinct roots, so it is not satisfied for 2/3 of the residues modulo p. Therefore, each of those non-roots is a value of u for which p must divide u2k + uk + 1. Also, since more than half of those residues qualify, we can choose an odd u, and then a = (u-1)/2 and b = (u+1)/2. With these values, p divides a2 + 3b2, which completes the proof of Lemma 1. LEMMA 2: If N is an integer of the form a2 + 3b2, and if the prime p = c2 + 3d2 divides N, then there exist integers u,v such that N/p = u2 + 3v2 and the representation of N is given by evaluating the product (p)(N/p) = (u2 + 3v2)(c2 + 3d2) using Fibonacci's formula. PROOF: Since p divides N, it must divide Nd2 - pb2. Also, we have which shows that the prime p must divide either ad+bc or ad-bc. Now, we can also write Depending on whether p divides ad+bc or ad-bc, we can choose the sign in the above expression so that p divides the right-most term. Then, since it also divides Np, it must divide the first term on the right. Therefore, dividing the above expression for Np by p2, we have N/p = u2 + 3v2 where u,v are the integers given by again with the choice of sign such that p divides ad-+bc. Solving these two equations for a and b gives This shows that the representation of N is given by applying Fibonacci's formula to multiply (p)(N/p), which completes the proof of Lemma 2. LEMMA 3: If we let [n\p] equal +1 or -1 accordingly as n is or is not a square (mod p), and if m,n are residues co-prime to p, then [mn\p] = [m\p][n\p]. PROOF: If m,n are both squares (mod p), then obviously mn is also a square. Also, if one of m,n is a square and the other is not, then it follows that their product mn is not, because if m=x2 and mn=y2 we would have n = (y/x)2, contrary to assumption that n is not square. The leaves only the case when neither m nor n is a square. To resolve this case, note that the non-zero multiplication table (modulo p) has unique inverse, so each non-zero residue appears in row and column precisely once. Also, since x2=y2 (mod p) implies (x-y)(x+y) mod p, it's clear that the squares of the residues 1 through (p-1)/2 are all distinct, and respectively equal to the squares of the residues (p+1)/2 to p-1. Therefore, the squares and non-squares each make up exactly half the non-zero residues. Also, each residue appears p-1 times in the table, so if fill in all the products of two squares, and all the products of a square and a non-square, we are left only with squares, which must be placed in the remaining openings, the products of two non-squares. Therefore [mn\p] = [m\p][n\p], completing the proof of Lemma 3. LEMMA 4: If the integer N is representable in the form a2 + 3b2 with (a,3b)=1, then the only odd prime factors of N are of the form p = 3k+1. PROOF: If N was divisible by a prime p, then we have a2 = -3b2 (mod p), which implies that (-3) is a square modulo p. It's easy to show that [-1\p] = (-1)(p-1)/2, and by quadratic reciprocity we also have [3\p] = [p\3](-1)(p-1)/2. From Lemma 3 and quadratic reciprocity it follows that [-3\p] = [-1\p][3\p] = [p\3]. Thus any number of the form a2 + 3b2 with (a,3b)=1 is divisible by only primes of the form 3k+1, which completes the proof of Lemma 4. Notes: 1. It's possible to avoid the use of full quadratic reciprocity here, but I wonder if Fermat might have just assumed it? 2. If a2 + 3b2 with (a,b)=1 is even, then a,b are odd, in which case either a+b or a-b must be divisible by 4. With that choice of sign we can set B=aħb and A=∓3b and then we have A2 + 3B2 = [a2+3b2]/4. Repeating if necessary, we can factor out all powers of 2, leaving an odd proper representation. LEMMA 5: Every prime p of the form 3k+1 is expressible in the form u2 + 3v2 with (a,b)=1 in precisely one way. PROOF: By Lemma 1 we know that p divides some integer of the form a2 + 3b2. Also, by replacing a and b with their least magnitude residues modulo p, the result is still divisible by p, but now we are assured that a and b are each less than or equal to (p-1)/2, from which it follows that a2 + 3b2 is strictly less than p2. Therefore, all the prime divisors of a2 + 3b2 other than p are strictly smaller than p, and according to Lemma 4 all of those prime divisors are of the form 3k+1, and according to Lemma 3 they are all of the form u2 + 3v2. Therefore, we can apply Lemma 2 to each of these smaller prime divisors in turn, yielding a unique quotient of the form a2 + 3b2, until arriving at p. This completes the proof of Lemma 5. LEMMA 6: The general primitive solution in integers of the equation x2 + 3y2 = N3 for odd N is given by x = u(u2 - 9v2) and y = 3v(u2 - v2) where u,v are co-prime integers. PROOF: By Lemma 4 we know that N3 is a product of primes of the form 3k+1, each of which by Lemma 5 has a unique proper representation of the form a2 + 3b2. Hence by Lemma 2 we can factor x2 + 3y2 uniquely into a product of primes of this form, and the representation of N3 is given by applying the Fibonacci product formula. Also, it's easy to verify that Fibonacci multiplication is commutative, in the sense that the two representations given by AB are the same as the two given by BA. Also, we can verify that Fibonacci multiplication is associative, i.e., (AB)C = A(BC), by noting the results where σ1 and σ2 are signed units (+1 or -1) according to our choices of signs in the two applications of the Fibonacci identity. Since both components are squared, we need consider only the magnitudes of the components, so we can multiply each term of the second component by σ1σ2 and write the two components as shown below Notice that the rows transpose (a,b), (c,d), and (e,f), so they have the same symmetry, and if we define σ3 = -σ1σ2 we have the three-way symmetry Consequently, the set of proper representations given by the Fibonacci product of three proper representations is the same, regardless of the order in which the product is evaluated. Furthermore, the number of distinct proper representations of a number equals 2k-1 where k is the number of distinct prime divisors, because we have two proper choices of sign when multiplying two distinct factors (whereas we have no proper choices when multiplying powers of a single prime). Since the number of distinct prime divisors of N is the same as the number of distinct prime divisors of N3, we can produce all 2k representations of N3 as the cubes of the 2k representations of N. Thus, for some co-prime integers u,v we have not only but also expanding the left side by the Fibonacci formula (which gives a unique proper result when cubing a single representation) we have completing the proof of Lemma 6. Now (finally!) we can return to our original problem. Recall that on the assumption of the existence of integers x,y,z such that x3 + y3 = z3, and assuming first that z is not divisible by 3, we had shown (see equations 2 and 3) the existence of integers m,n and co-prime integers u,v such that where n is odd. It follows from Lemma 6 that u and v can be expressed in terms of integers r,s as follows Also, since v is odd and u is even, we must have r even and s odd. Further, since u = 4m3, it's clear that r is 4 times a cube, and both r-3s and r+3s are cubes. Thus we have and therefore from 2r = (r-3s) + (r+3s) we have which is a solution of the original equation in strictly smaller integers. However, by applying the same argument to this new solution we can construct a strictly smaller solution, and so on, ad infinitum. This is clearly impossible, since there must be some absolutely smallest integer solution. Consequently, by Fermat's principle of infinite descent, we see that solutions with z not divisible by 3 are impossible. For the second case, suppose z is a multiple of 3. It follows that u is a multiple of 3, and v is not. In this case we cannot say that 2u is co-prime to u2 + 3v2, because both are divisible by 3, but if we factor a 3 out of the quantity in parentheses in (1) we have so now 6u is co-prime to the quantity in brackets, and so both factors are cubes, which implies From Lemma 6 we have co-prime integers r,s with s even, such that which implies so we have and therefore Since 2s + (r-s) = (r+s) we have so again we have a solution in smaller integers, and by the principle of infinite descent, this is impossible. Consequently, we have proven the result THEOREM: The equation x3 + y3 = z3 has no solution in non-zero integers. Return to MathPages Main Menu