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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