In today's blog, I will continue the proof for Fermat's Last Theorem n=7.
The details for today are based on Paulo Ribenboim's Fermat's Last Theorem for Amateurs.
Lemma 1: Given u2 = s4 + 6t2s2 - t4/7 where s,t are integers and gcd(s,t)=1, we can conclude that there exists an integer v such that:
(a) t = 7ev
(b) 7 doesn't divide v
(c) gcd(s2 + 3*72ev2 + u,s2 + 3*72ev2 - u) is a power of 2 or 1.
(1) s,t are integers so 7s4 + 42t2s2 - t4 is an integer.
(2) But this means that 7u2 is an integer which means that u2 is an integer. If u2 were not an integer then it u = would have to equal some number of √7 which is impossible since u is a rational number and √7 is not (see here for details).
(3) But from (a), we also note that 7 must divide t4 which means from Euclid's Generalized Lemma, 7 must divide t.
(4) If e is the number of times that 7 divides t, then there exists v such that:
t = 7ev and 7 doesn't divide v.
(5) Inserting this into the value of u2 gives us:
u2 = s4 + 6*(72e)v2s2 - 74e-1v4 = (s2 + 3*72ev2)2 - 64*74e-1v4
(6) Subtracting u2 from both sides and adding 64*74e-1v4 to both sides gives us:
(s2 + 3*72ev2)2 - u2 = (s2 + 3*72ev2 + u)(s2 + 3*72ev2 - u) = 64*74e-1v4
(7) Now, we can show that gcd( s2 + 3*72ev2 + u , s2 + 3*72ev2 - u) = 2f since:
(a) Assume that a prime p divides both and it is not 2.
(b) Then p divides u [since 2u = s2 + 3*72ev2 + u -( s2 + 3*72ev2 - u) and p doesn't divide 2.]
(c) We also know that p divides s2 + 3*72ev2 [since 2(s2 + 3*72ev2) = s2 + 3*72ev2 + u + (s2 + 3*72ev2 - u) and p doesn't divide 2]
(d) We know that p ≠ 7 since then 7 would divide s but this is impossible since 7 divides t and gcd(s,t)=1.
(e) We know from #6, that p divides 64*74e-1v4
(f) This means that p must divide v (since we are assuming it is not 2 and we know it is not 7)
(g) Going back to (c), it must also divide s.
(h) But this is impossible since it would then also divide t (since t = 7ev) and we know that gcd(s,t)=1.
(i) So we have a contradiction and we reject our assumption (7a).