In the three centuries since its proposal, Fermat's Last Theorem has led to the development an enormous amount of mathematics, both good and bad. On the good side, it would not be too far-fetched to assert that much of modern algebraic number theory has sprung from attempts to prove what Fermat is alleged to have written in the margin of his copy of Diophantus in 1637. However, because of the problem's simple statement it has also generated a mountain of material which as best was wrong and at worst not even classifiable as mathematics.

For the past seven years, James Harris has been a regular fixture on the Usenet newsgroup, sci.math. James, an amateur mathematician, has produced scores, if not hundreds of self-proclaimed elementary "proofs" of Fermat's Last Theorem, none of which, to date, have withstood serious scrutiny. Due to James's unfamiliarity with mathematical exposition, the work available on his main and "area one" web sites and his contributions to the newsgroup have often been difficult to understand and have been augmented and clarified by many others over the years. This page is my attempt to gather together what has been said by and to Mr. Harris about FLT and present it in a style which meets at least minimal standards for clarity.

Rick Decker

We begin by establishing four preliminary results.

Lemma 1. If p is an odd prime, p(r + s)rs | (r + s)^p - r^p - s^p. Denote by H_p the polynomial

H_p = ((r + s)^p - r^p - s^p) / (p(r + s)rs).

Then H_p(r, s) = G_p(r + s, rs), where G_p is a polynomial with integer coefficients.

Proof. We observe that each term of (r + s)^p - r^p - s^p has as its coefficient an order-p binomial coefficient, and it is well known that if p is prime, all of these will be divisible by p. It is easy to verify that both rs and r + s divide (r + s)^p - r^p - s^p. To establish the last statement of the Lemma, we observe that H is a symmetric polynomial and that any such polynomial may be expressed in terms of the elementary symmetric functions, which in this 2-variable case are r + s and rs. [Note 1.]

Lemma 2. Let p be an odd prime and x, y, z be integers satisfying x^p + y^p = z^p and let v be an arbitrary integer. Then

(v^p + 1)z^2p - pv(xy)² z² G_p(vz², (xy)²) - 2 (xy)^p = 0 mod (x² + y² + vz²),

where G_p is defined in Lemma 1.

Proof. For any integer v we have, trivially,

x² + y² = -vz² mod (x² + y² + vz²).

Raising both sides to the power p gives

(x² + y²)^p = -(vz²)^p mod (x² + y² + vz²).

Expanding the left side and using Lemma 1 gives

x^2p + p(xy)²(x² + y²)H_p(x², y²) + y^2p = -(vz²)^p mod (x² + y² + vz²).

Since we assume that x, y, and z satisfy x^p + y^p = z^p we can use (x^p + y^p)²= z^2p to eliminate x^2p and y^2p in the congruence above, yielding

(v^p + 1)z^2p - p(xy)²(x² + y²)H_p(x², y²) - 2(xy)^p = 0 mod (x² + y² + vz²).

Again referring to Lemma 1, we see that H_p(x², y²) may be expressed as a polynomial in x² + y² and (xy)², so modulo x² + y² + vz² may be denoted by G_p(vz², (xy)²), yielding the desired result.

Lemma 3. Let p be an odd prime. Let x, y, z be pairwise coprime positive integers satisfying x^p + y^p = z^p. Then

1. One of z - y, z - x, or x + y has a prime factor q ≠ p.
2. Without loss of generality, suppose q | z - y. Then q | x, and if x = q^ju, with q not dividing u, then q^jp | z - y.

Proof. First, it is easy to show that no two of z - y, z - x, x + y can be 1. If, for example, z - y = z - x = 1, then we would have x = y, contradicting the relative primality of x and y (since we obviously couldn't have x = y = 1). The other two cases are handled similarly. It is also easy to show that no two of z - y, z - x, x + y can be divisible by p If, for example, z = y mod p and z = x mod p, then, applying Fermat's little theorem to x^p + y^p = z^p, we would have x + y = z mod p, so we would conclude that x = 0 mod p and y = 0 mod p, again contradicting the relative primality of x and y. The remaining two cases are handled similarly. Combining these two results, we see that at least one of z - y, z - x, or x + y must have a prime factor q unequal to p.

Suppose that q | z - y. Then, since

x^p = z^p - y^p = (z - y)(z^p-1 + z^p-2y + ... + zy^p-2 + y^p-1)

it is clear that z - y | x^p, so q | x. It is an old result that for prime p, z - y and (z^p - y^p) / (z - y) have no factors in common except for p, so if x = q^ju, with q not dividing u, then by inspecting the orders of q in the terms x, x^p, and z - y, we conclude that q^jp | z - y.

Lemma 4. Let R be a commutative ring containing the integers. Suppose that q is a rational prime and t is an integer not divisible by q. Suppose also that q | A(t + qC) in R, for some A and C in R. Then, q | A in R.

Proof. Since q and t are relatively prime in Z, there exist integers r and s such that rq + st = 1. Since q | A(t + qC) in R, we have q | At in R and obviously q | Arq in R, so q | Ats + Arq = A(ts + rq) in R and hence q | A in R, as required.

With the preliminaries out of the way, we establish the two congruences that constitute what James calls "Area One."

1.1. Divisibility results

Let p be an odd prime and let x, y, z be pairwise coprime nonzero integers satisfying x^p + y^p = z^p. From Lemma 3, without loss of generality we may suppose that q ≠ p is a prime factor of x and z - y. [Note 2.] Then q must be a factor of x² + y² - z² as well. Let x = q^ju, where q does not divide u.

Theorem 1. With the assumptions above, for any positive integer n there exists an integer m for which

Proof. Observe that from Lemma 3, q^pj | z - y, so we have that q^2j divides y² - z². Since q does not divide z (otherwise we would contradict the relative primality of x and z), we can solve the congruence [1] for m, as required.

From this, we have an immediate consequence,

Corollary 1. For any integer k > 2j, there exists m such that if v = -1 + m^2j then

1.2. The polynomial F_{p, v}(z², xy)

Denote the left hand side of the congruence of Lemma 1 by F = F_{p, v}(z², xy), so that

For any fixed values of p and v, with v ≠ -1 we may consider F as a degree p polynomial in the indeterminants z² and xy, so we may factor F into terms linear in z² and xy, over the complex numbers:

Equating coefficients, we observe that a₁ ... a_p = (v^p + 1). By appropriate multiplication and division we may write [2] as

In what follows, we will write the factorization of F as

where c₁ = c₂ = (v + 1)^1/2, c₃ = (v^p + 1) / (v + 1), and c_i = 1 for 3 < i ≤ p, and the d values are defined as above.

Now from the polynomial identity [3], we return to regarding F_{p, v} as an integer, using the original values for x, y, and z, namely a Fermat counterexample for a specific p.

1.3. Divisibility in the overring

Let k > 2j be an arbitrary integer and fix v to be -1 + mq^2j where m is as in Corollary 1. Define the ring R = Z[m^1/2, d₁, d₂, ... , d_p] for the ds in [3] resulting from this choice of v. [Note 3.] So, since q^k | x² + y² + vz² from Corollary 1 and x² + y² + vz² | F_{p, v} from Lemma 2 we have (considering all subsequent congruences in the context of R, which we shall denote by =_R for emphasis),

We then have

Theorem 2. Suppose F_{p, v} = (c₁z² + d₁xy)(c₂z² + d₂xy) ... (c_pz² + d₁xy) as in [3]. Then, for any integer n > 0 there exists an integer m such that with v = -1 + m^2j,

Proof. Using the values of c_i established in [3], we have

and dividing out q^j from each of the first two factors yields

Denote by A the product of the first 2 factors in [4] and by B the product of the remaining p - 2 factors, we have, after expanding B,

and

Letting t = ((v^p + 1) / (v + 1))z^{2(p - 2)} (and observing that in Z, q does not divide the integer t) we can rewrite [4] as

and by repeated applications of Lemma 4 we conclude that

Writing k - 2j as n establishes our result.

1.4. Consequences

For convenience, we recall our terminology.

• p is an odd prime.
• x, y, z are pairwise coprime positive integers satisfying x^p + y^p = z^p.
• q ≠ p is a prime which, without loss of generality, we may suppose divides x and y - z.
• x = q^ju, with q not dividing u.
• For n > 0, m = m(n) is a solution to mz² = -(u² + (y² - z²) / q^2j) mod qⁿ.
• v = -1 + mq^2j. Observe that v ≠ -1.
• F_{p, v} is factored as (c₁z² + d₁xy) . . . (c_pz² + d_pxy), with c₁ = c₂ = (v + 1)^1/2, c₃ = (v^p + 1) / (v + 1), and c_i = 1 for 3 < i ≤ p.
• R = R(m) = Z[m^1/2, d₁, ... , d_p].
• S(m) = u² + (y² - z²) / q^2j + mz².
• T(m) = mz⁴ + m^1/2(d₁ + d₂)z²uy + d₁d₂u²y².

Finally, recall that by Theorems 1 and 2, for any n > 0 we can find an m = m(n) so that

and in R = R(m)

Note that changing the exponent r in the congruence modulo q^r that defines m will in general change m, which will change v, which will change F_{p, v}, which will change the c and d values, which will give rise to a different ring R. However, if we have defined m(r), then in any congruence modulo q^s for s ≥ r, m(s) will serve equally well as a solution for m(r) with no change in the ring R.

The next result and its corollaries establishes some further (and somewhat surprising) relations among the d values and the nature of q in the ring R.

Theorem 3. Let m = m(j). Then, in the ring R = R(m) we will have d₁ + d₂ =_R 0 mod q^j.

Proof. Let F_{p, v} be the polynomial defined in [3] for v = -1 + mq^2j. Observe that in F_{p, v} = (v^p + 1)z^2j - pvz²(xy)²G_p(z², xy) - 2(xy)^p, G_p is a degree-(p - 3) polynomial in z², so in F_{p, v} the (z²)^p-1 term will have coefficient 0. If we expand the right side of the factorization of F_{p, v} and equate coefficients we will have

Using the values for the cs gives us

So, since v ≠ -1, we have

Denoting the factor on the right by D, we may write this as

Thus, in the ring R we have

Since (v^p + 1) / (v + 1) is an integer not divisible by q it has a multiplicative inverse in Z modulo q^j, so we may write

Finally, recall that v = -1 + mq^2j so the factor (v + 1)^1/2 on the right will be divisible (in R) by q^j, and thus

Corollary 2. With R defined as above, d₁d₂ =_R 1 mod q^j.

Proof. Since we have just established d₁ + d₂ =_R 0 mod q^j and we know from Theorem 2 that

we find that

Now since y = z mod q^pj by Lemma 3, we may set y = z above and since neither is zero modulo q we may multiply both sides by (z²)^-1 modulo q in Z, giving us

Recall that m is defined by m = -(u² + (y² - z²) / q^2j) / z² mod q^j and observe that q^j divides y² - z²) / q^2j so we have m = -u² / z² mod q^j. Substituting this into [8] yields

and since u is not divisible by q in Z we may multiply both sides of the above congruence by a multiplicative inverse of u mod q^j to obtain the desired result.

In his so-called "Area 2," James now proceeds to use S(m) and T(m) to get an expression not apparently involving m. Recall that for any n > 0 we may find a rational integer m = m(n) so that

and in R = R(m),

For simplicity, let A = u² + (y² - z²) / q^2j. Multiplying [9] by z² then gives us

Substituting into [10] and rearranging yields

Squaring both sides of the congruence above, we have

Expanding the right hand side and using [11] again, we have

So, finally, we have, after simplifying,

Which, we note in passing, may be written

It appears that we have eliminated m from our expressions, but nothing could be further from the truth. While it is certainly true that q, u, y, and z (and hence A) do not depend on m, certainly the d values and the underlying ring R all depend on m(n). What we have, then, are a family of congruences that posit some mildly interesting divisibility results in different rings. What we certainly do not have at this stage is anything that establishes a contradiction.

Credits

This explication of James Harris's most recent attempt at a proof of Fermat's Last Theorem using elementary methods is the distillation of the work of many people on the sci.math newsgroup. Since this discussion has taken place over the span of seven years, it is highly likely that I have forgotten to credit some of the contributors, but much of what appears here is due to comments by (in no particular order) David Libert, John Rickard, Arturo Magidin, Nora Baron, "The Scarlet Manuka," Dik T. Winter, John Roberts-Jones, Keith Ramsay, and I'm certain, many others. If I've left off your name and you'd like credit, let me know and I'll be happy to include you and, of course, I'll be happy to entertain any suggestions for modifications to this page.
Back to the paper

Notes

1. For example, if we let p₁ = r + s and p₂ = rs, then
   H₃(r, s) = G₃(p₁, p₂) = 1
   H₅(r, s) = G₅(p₁, p₂) = (p₁)² - p₂
   H₇(r, s) = G₇(p₁, p₂) = (p₁)⁴ - 2(p₁)²p₂ + (p₂)². Back to the paper

2. For the benefit of those people following the discussions with James, I should note that what James has been calling f I have renamed q for stylistic reasons. In addition, in the factorization [2] I have used d_i for what James has been calling b_i. Back to the paper

3. In the original argument, R was taken to be a ring containing the integers, the square roots of the integers, and the ds. I have reduced the ring by excluding all the square roots except that of m (and any arising from the ds). Note here that there is a dependency among the ds from [3], since the product d₁ ... d_p = -2. [In the two years since this page was produced, James has admitted to problems with his approach and attempted to recast his arguments using the ring of algebraic integers for R.] Back to the paper

Last modified 2/24/04