@JohnsonFido @capobianco_slv @GWOMaths Got it.
Given
a² - d⋅b² = ±4
we have, equivalently
(½a)² - d⋅(½b)² = ±1
Evaluating:
(½a - √d⋅½b)³
= a⋅½(db²±1) - b⋅½(a²∓1)⋅√d
where both of
(db²±1)
and
(a²∓1)
are always even for d, a, and b odd.
But for d ≅ 1 (mod 4)
then a and b are always the same parity.
Given
a² - d⋅b² = ±4
we have, equivalently
(½a)² - d⋅(½b)² = ±1
Evaluating:
(½a - √d⋅½b)³
= a⋅½(db²±1) - b⋅½(a²∓1)⋅√d
where both of
(db²±1)
and
(a²∓1)
are always even for d, a, and b odd.
But for d ≅ 1 (mod 4)
then a and b are always the same parity.
@JohnsonFido @capobianco_slv @GWOMaths Thus
x = a⋅½(db²±1)
y = b⋅½(a²∓1)
are positive integer solutions to
x² - d⋅y² = ±1
whenever
a² - d⋅b² = ±4
for positive integers a, b.
I believe this is only relevant for d ≅ 1 (mod 4) even though that condition doesn't appear in the calculation.
x = a⋅½(db²±1)
y = b⋅½(a²∓1)
are positive integer solutions to
x² - d⋅y² = ±1
whenever
a² - d⋅b² = ±4
for positive integers a, b.
I believe this is only relevant for d ≅ 1 (mod 4) even though that condition doesn't appear in the calculation.
@JohnsonFido @capobianco_slv @GWOMaths Derivation:
(±1)³
= (½a - ½b⋅√d)³
= (½a - ½b⋅√d) ⋅ ¼⋅(a² + d⋅b² - 2ab⋅√d)
= ⅛⋅(a³+da⋅b²+2⋅da⋅b²) - ⅛⋅(a²⋅b+d⋅b³+2a²⋅b)⋅√d
= ⅛⋅(a³+3da⋅b²)- ⅛⋅(d⋅b³+3⋅a²⋅b)⋅√d
= ⅛⋅(4da⋅b² ± 4) - b⋅⅛⋅(4⋅a² ∓ 4)⋅√d
= a⋅½(db²±1) - b⋅½(a²∓1)⋅√d
(±1)³
= (½a - ½b⋅√d)³
= (½a - ½b⋅√d) ⋅ ¼⋅(a² + d⋅b² - 2ab⋅√d)
= ⅛⋅(a³+da⋅b²+2⋅da⋅b²) - ⅛⋅(a²⋅b+d⋅b³+2a²⋅b)⋅√d
= ⅛⋅(a³+3da⋅b²)- ⅛⋅(d⋅b³+3⋅a²⋅b)⋅√d
= ⅛⋅(4da⋅b² ± 4) - b⋅⅛⋅(4⋅a² ∓ 4)⋅√d
= a⋅½(db²±1) - b⋅½(a²∓1)⋅√d
• • •
Missing some Tweet in this thread? You can try to
force a refresh
