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The following Diophantine equation is defined for positive integers a, b, c, k and
exponent n, where 1 < a < b < c, n >= 2, k >= 0, GCD(a,b)=GCD(a,c)=GCD(b,c)=1

1) c^n = k.a^n.b^n + a^n + b^n

Solution If q_a and q_b denote ‘unity roots’ mod a^n and mod b^n respectively, defined by

2)      q_a^n = 1 mod a^n
3)      q_b^n = 1 mod b^n

then equation 1 has integer solutions for c (and consequently k), parameterised
by integers s and t, as given by

4)      c = s.a^n + q_a.b
5)      c = t.b^n + q_b.a

where integers s and t are solutions to the following linear Diophantine
equation obtained by equating 4 and 5

6) s.a^n - t.b^n = q_b.a - q_a.b

Example a = 2, b = 3, n = 3, i.e. the cubic case, q_a = 1, q_b = 19,

we find that a particular solution for s and t is

s = 1 and t = -1

so that equation 4 then gives c = 11 since

11 = 1.2^3 + 1.3

and equation 5 also gives c = 11 since

11 = -1.3^3 + 19.2

hence equation 1 becomes

11^3 = 6.2^3.3^3 + 2^3 + 3^3

whereby the k = 6 value was obtained by calculating

6 = (11^3 - 2^3 - 3^2)/(2^3.3^3)

Commentary The example above is the only case known where the equality holds since 11 = 2^3 + 3 all other cases seem to show c >> a^n + b and, consequently, k >> 0.

Because the equation has solutions it is considered more tractable to a simple proof than FLT. If the MFLT conjecture is true it implies k > 1 and therefore a proof of the MFLT conjecture is also a proof of FLT.

A computer analysis has found no counter-examples to the MFLT conjecture (yet) but neither has it been proven outright. In particular, the proof of the case where s < 0, t < 0, q_a > 1, q_b > 1 currently eludes the author!
R J Miller, Dec. 2013 (last revised Nov 2006). Back to home page.

Equation 1 is termed ‘The Modified FLT Equation’ abbreviated to simply ‘MFLT’. Unlike FLT it has an analytic solution (below) and is therefore tractable to analysis.