8 Mar, 6 tweets, 3 min read
@GWOMaths Two important concepts are:

i) That an expression such as
(5 + √21)
CAN be a perfect square; and

ii) That the resultant square root looks like
(a + b⋅√21)

Then set
(5 + √21)
= (a + b⋅√21)²
= (a² + 21⋅b²) + (2ab √21)

Hence
(1) 5 = a² + 21⋅b²
(2) -1 = 2ab

1/
@GWOMaths From (2):
b = -1 / 2a
and
b² = 1 / 4⋅a².

Substituting into (1);
then rearranging and multiplying through by 4⋅a²:
4⋅a⁴ - 20⋅a² + 21 = 0.

Note that
4⋅21 = 2⋅2⋅3⋅7 = (2⋅3) ⋅ (2⋅7) = 6⋅14
and thus
4⋅a⁴ - 20⋅a² + 21 = (2⋅a² - 3) ⋅ (2⋅a² - 7)

2/
@GWOMaths Hence:
a² = 3/2 or 7/2
a = ±√(3/2) or ±√(7/2)
b = -1 / 2a = ∓1/√6 or ∓1/√14
(a + b⋅√21) = ±( √(3/2) - √(7/2) ) or ±( √(7/2) - √(3/2) )

with those two latter expressions just the sign reversal of each other.

We desire the POSITIVE square root.
@GWOMaths Therefore:
√( 5 - 21 ) = ( √(7/2) - √(3/2) ) = (√7 - √3) / √2.

Substituting back into the problem:
1 / (√x + √3) = ½ ⋅ √½ ⋅ ( (√7 - √3) / √2 )
= ½ ⋅ ½ ⋅ (√7 - √3)
= ¼ ⋅ (√7 - √3).
@GWOMaths Multiply the left-hand side by
(√x - √3) / (√x - √3)
and substituting yields:
1 / (√x + √3) = ¼⋅ (√x - √3)

and thus that
¼⋅ (√x - √3) = ¼ ⋅ (√7 - √3).

A little algebra quickly gives that
x = 7.

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# More from @pgeerkens

10 Feb
@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.
@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.
17 Jan
@johnrobb We have abandoned competence,
in the futile search for universal expertise.

"What?" I hear you say.
Clearly definitions and explanations are in order.

1/
@johnrobb Expertise, certainly as regards this discussion,
but perhaps also universally, should be regarded
as the knowledge of a very great deal - about very
very little.

2/
@johnrobb To become an expert in a field requires
years - nay, decades - of diligent study of that
solitary field. In diving deeper, the pool in turn
becomes ever smaller: a diving pool rather than a
swimming pool. Nothing is free.

3/
15 Jan
@GWOMaths One might say that physicists study the symmetry of nature, while mathematicians study the nature of symmetry.

1/
@GWOMaths Observing symmetry in nature, such as noting the similarity between the symmetries of a snowflake and a hexagon, is readily comprehensible. What does it mean then to study "the nature of symmetry"?

2/
@GWOMaths Mathematicians define a "group", G, as a set of elements {a,b,c, ...} with a binary operation ⊡ and a distinguished element e (the identity of G) satisfying these specific properties:

3/
2 Dec 20
@pnjaban None of the above:
Field-trained former military paramedic.

I don't want anyone not trained specifically in emergency medicine anywhere near; and none of those civilian trained E.R. either, cause they're trained in CYA first, and I want someone who's serious about saving my life
@pnjaban When my Dad suffered acute hemechromatosis at ~34,
for which the ONLY treatment is *bleeding*,
and the only options "a whole lot" and "a whole lot more",
the attending physician was too embarrassed to prescribe the correct treatment.
@pnjaban The nurse (Irish of course, because they understand hemechromatosis) told me Mom:

Mrs. Geerkens:
unless we remove him from this hospital immediately.
I'll help.

The nurse lost her job;
but my Dad lived another 45 years.
2 Dec 20
From Albert H. Beiler's
Recreations in the Theory of Numbers:

47² = ?
96² = ?
113² = ?
179² = ?

47² = (47 + 3) ⋅ (47 - 3) + 3² = 50 ⋅ 44 + 9 = 2209

96² = (96+4) ⋅ (96-4) + 4² = 100 ⋅ 92 + 16 = 9216

113² = (113+3) ⋅ (113-3) + 3² = 116⋅110 + 9
= 12760 + 9 = 12,769

179² = (179+21) ⋅ (179-21) + 21²
= 200 ⋅ 158 + (20² + 20 + 21)
= 31600 + 441 = 32041
The trick is from noting that from difference of squares
a² - b² = (a + b) ⋅ (a - b)
one can obtain by rearrangement
a² = (a + b) ⋅ (a - b) + b².
22 Nov 20
@GWOMaths Hint follows
@GWOMaths Rochambeau is just a fancy name for Rock-Paper-Scissors.
@GWOMaths If you're having difficulty visualizing this,
consider the situation after Game 1 as the baseline.