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MATHS SUPPLEMENT to PROSPERO 24-4
Page 3 first* The method is similar to that used in STEP 10.
Page 4 second *: From the 43rd prime number onwards (191) theta is < 1degree. So in the vast majority of prime number cases p, theta is tiny and A is very close indeed to the real axis.
Page 4 third* There is a typo in equation (16). For 'P' read 'B'. The next line takes it that x in the numerator of (16) must cancel with a factor x in the denominator; so the factor x on the RHS of (16) has been discounted.** (Otherwise A has the factor p, which is a contradiction (see below).) [NOTE when (16) is referred-to as an expression, it is the RHS of (16) which applies.]
SOME LEMMAS
x has the factor 2:
Z = A + B - x. When Z is even, A, B are odd, A+B is even and hence x is even. Similarly A = Z- B+z. When A is even Z-B is even and x is even. Similarly when B is even.
x has the factor p:
Z = (A+B) -x, so z = (A+B)^p - pZ(A+B)xW - x^p, where W is the inner tray. This also equals a + 2pABZV + b.
So pAB(A+B)T - pZ(A+B)xW - 2pABZV = x^p, where T is the inner tray of (A+B)^p. The LHS has a factor p, so the RHS must have a factor p. Hence x has a factor p.
V has the factor p:
In case (1) v has the factor p^(p-1). v = T + powers of h, so T has the factor p in this case. So three lines above the term 2pABZV is the only one which does not appear to have the factor p^2. We know that in case (1) none of A, B, Z have the factor p. So V must have the factor p.
T' has a factor p:
T' = terms with a factor x plus t, where t = the inner tray of (A+B)^p. t, we know, must have a factor V and, in case (1), an extra factor p.
2V-T' has a factor p:
because both V and T' do. We can focus on it by subtracting z as (A+ B')^p from z as a+b+2h. Let x=pX. The result after dividing by p: ABZ(2V-T')/p = -XB^(p-1) (mod p).
Also: the denominator will have a factor p^2 iff BZ(2V-T')/p -XBT" = 0 (mod p). We know T" = B^(p-3) + x terms.
This means that the iff condition is ABZ(2V-T')/p = AXB.B^(p-3) (mod p).
So the denominator of (16) will have a factor p^2 iff -XB^(p-1) - AXB^(p-2) = 0 (mod p) or B+A =0 (mod p) which entails that Z has the factor p, contrary to hypothesis. This is a contradiction. [Note that z=a+b + 2pABZV, so Z=A+B (mod p).
Incidentally the symmetry of A, B and -Z explained in Step 6 implies that A has the factor p. This also implies that B has a factor p and the numerator in (16) has a factor p^3 while the dnominator has a factor p^2. These are all aspects of the same contradiction.

**The denominator of (16) consists of terms in x plus BZ(2V - T'). If this expression is divisible by x, the reasoning in the supplement applies. If it has no common factor with x, then the denominator of (16) has no part of factor x, and the x in the numerator of (16) must consist entirely of subfactors of A, which is a contradiction because mutatis mutandis it also consists of subfactors of b. The intermediate case remains: but it is superseded by the reasoning above.


Page 4 fourth-fifth* are in the same case.

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