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10.12.10

Some "Technicalities"

Assume P == 1 (mod 4) and Q == 1 (mod 4) are Euler primes (with Q <= P) such that



L(Q) < U(P)


where:


L(x) = (3x^2 - 4x + 2)/(x(x - 1))

U(x) = (3x ^2 + 2x + 1)/(x(x + 1))


Note that:


L(x) = 3 - [(x - 2)/(x(x - 1))]

U(x) = 3 - [(x - 1)/(x(x + 1))]


Hence,


L(x) + [(x - 2)/(x(x - 1))] = U(x) + [(x - 1)/(x(x + 1))] = 3


Then since L(x) and U(x) are increasing functions of x, then assuming Q <= P:


L(Q) <= L(P)

U(Q) <= U(P)


But we also know that:


L(E) < U(E) for all Euler primes E

( i.e. L(E) < I(E^k) + [2/(I(E^k)] <= U(E) )


We actually have the "stronger" inequality:


L(x) < U(x) for all x in [a, b]


for some suitable closed interval [a, b] to be determined in a while.


Consider the auxiliary function:


A(x) = L(x) - U(x) for all x in [a, b]


Let us now compute:


A(a) = L(a) - U(a) < 0

A(b) = L(b) - U(b) < 0


In particular:


A(y) = L(y) - U(y) < 0, for all y in the open interval (a, b)


First, assume:

{x_1} = a = b = Q


Then A({x_1}) = A(Q) = L(Q) - U(Q) < 0 (where {x_1} = Q)


Next, assume:


{x_2} = a = b = P.


Then A({x_2}) = A(P) = L(P) - U(P) < 0 (where {x_2} = P)


If Q = P, then A(Q) = A(P). (Note that A(x) is defined for all x in the closed interval [Q, P].)

{  If A(Q) = A(P), does it follow that Q = P?
YES.  (The WolframAlpha verification is hyperlinked here.)  }


This implies that, A(Q) - A(P) = 0 if and only if Q = P.


But:

A(Q) = L(Q) - U(Q)

AND

A(P) = L(P) - U(P).


Consequently:

0 = A(Q) - A(P) = [L(Q) - L(P)] + [U(P) - U(Q)]  if and only if Q = P.


But L(Q) <= L(P) implies that [L(P) - L(Q)] >= 0.
(This means that [L(Q) - L(P)] <= 0.)

Likewise, U(Q) <= U(P) implies that [U(Q) - U(P)] <= 0.
(This means that [U(P) - U(Q)] >= 0.)


But, then again:


0 = [L(Q) - L(P)] + [U(P) - U(Q)]   if and only if Q = P


Hence you have:


(*) [L(Q) - L(P)] <= 0
AND [U(Q) - U(P)] <= 0 for all possible pairs (Q, P) (*)


Now, consider the following (second and third) auxiliary functions:

B(x,y) = L(x) - L(y) for all possible pairs (x, y) in (a, b)

C(x,y) = U(x) - U(y) for all possible pairs (x, y) in (a, b)

{  Note that the open interval (a, b) here is given by (Q, P).  }


Then (*) implies that:

(**) B(Q,P) = [L(Q) - L(P)] <= 0
AND C(Q,P) = [U(Q) - U(P)] <= 0
for all possible pairs (P,Q) (**)


Let us now try to compute the exact formulas for B(Q,P)
and C(Q,P).

B(P,Q) = [L(Q) - L(P)]

L(x) = 3 - [(x - 2)/(x(x - 1))]

L(Q) - L(P) = { 3 - [(Q - 2)/(Q(Q - 1))] } - { 3 - [(P - 2)/(P(P - 1))] }

= [(P - 2)/(P(P - 1))] - [(Q - 2)/(Q(Q - 1))]

= [Nume(L, P, Q)]/[Deno(L, P, Q)]

where

[Nume(L, P, Q)] = [{Q(Q - 1)}(P - 2) - (Q - 2){P(P - 1)}]

and

[Deno(L, P, Q)] = [(P(P - 1))(Q(Q - 1))]


[Nume(L, P, Q)] = [{Q(Q - 1)}(P - 2) - (Q - 2){P(P - 1)}]

= (Q - P)(QP - 2Q - 2P + 2)

= (Q - P)[Q(P - 2) - 2(P - 1)]

= (Q - P)[Q(P - 2) - 2(P - 2) + (-4 + 2)]

= (Q - P)[(Q - 2)(P - 2) - 2]

= (Q - P)(Q - 2)(P - 2) - 2(Q - P)


Thus,

[Nume(L, P, Q)] = (Q - P)[(Q - 2)(P - 2) - 2]
= (Q - P)(Q - 2)(P - 2) - 2(Q - P)

and

[Deno(L, P, Q)] = [(P(P - 1))(Q(Q - 1))]



C(Q,P) = [U(Q) - U(P)]

U(x) = 3 - [(x - 1)/(x(x + 1))]

U(Q) - U(P) = { 3 - [(Q - 1)/(Q(Q + 1))] } - { 3 - [(P - 1)/(P(P + 1))] }

= [(P - 1)/(P(P + 1))] - [(Q - 1)/(Q(Q + 1))]

= [Nume(U, P, Q)]/[Deno(U, P, Q)]

where [Nume(U, P, Q)] = [{Q(Q + 1)}(P - 1) - {P(P + 1)}(Q - 1)]

and [Deno(U, P, Q)] = [(P(P + 1))(Q(Q + 1))]

[Nume(U, P, Q)] = [{Q(Q + 1)}(P - 1) - {P(P + 1)}(Q - 1)]

[Nume(U, P, Q)] = (Q - P)[(Q - 1)(P - 1) - 2]


Now, compare B(Q,P) and C(Q,P):

{ [Nume(L, P, Q)] / [Deno(L, P, Q)] }

compared to

{ [Nume(U, P, Q)] / [Deno(U, P, Q)] }

which gives us:

[Nume(L, P, Q)][Deno(U, P, Q)]

compared to

[Nume(U, P, Q)][Deno(L, P, Q)]


Thus:

(Q - P)[(Q - 2)(P - 2) - 2][(P(P + 1))(Q(Q + 1))]

compared to

(Q - P)[(Q - 1)(P - 1) - 2][(P(P - 1))(Q(Q - 1))]


Consequently, assuming Q < P, we are essentially comparing:

D(Q,P) = [2 - (Q - 2)(P - 2)][(P + 1)(Q + 1)]

with

E(Q,P) = [2 - (Q - 1)(P - 1)][(P - 1)(Q - 1)]


Here are the computations from WolframAlpha:

Case 1:  D(Q,P) < E(Q,P)

Inequality Plot:


Integer Solution:  P = 0, Q = 0


Case 2:  D(Q, P) > E(Q,P)

Inequality Plot:


Integer Solutions:  P in {-8, -7, -6, -5, -4, -3, -2, 2, 3}

Case 3:  D(Q, P) = E(Q, P)

Equality Plot:


A natural question to ask at this point would be:  Are there any lattice points (P, Q) on this curve satisfying P = Q?
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