**
This Thread may be Removed Anytime!**

Twitter may remove this content at anytime! Save it as PDF for later use!

- Follow @ThreadReaderApp to mention us!
- From a Twitter thread mention us with a keyword "unroll"

`@threadreaderapp unroll`

Practice here first or read more on our help page!

@GWOMaths Spoiler - hint and solution follows

@GWOMaths Define Sₖ(c) as the sum of k consecutive integers with offset c thus:

Sₖ(c) = (c+1) + (c+2) + ... + (c+k-1) + (c+k)

= k⋅c + (1 + 2 + ... + (k-1) + k )

= k⋅c + ½k(k+1).

Sₖ(c) = (c+1) + (c+2) + ... + (c+k-1) + (c+k)

= k⋅c + (1 + 2 + ... + (k-1) + k )

= k⋅c + ½k(k+1).

@GWOMaths We desire to find which n ∈ Z⁺ with 50 ≤ n ≤100 can be expressed as Sₖ(c) for some k ≥ 3 and c ≥ 0.

Case 1 - k is odd:

Sₖ(c)

= k ⋅ (c + ½(k+1))

= k ⋅ ½(2c + k + 1)

where ½(2c + k + 1) is integral.

Case 2 - k is even:

Sₖ(c)

= ½k ⋅ (2c + k+1)

where ½k is integral.

Case 1 - k is odd:

Sₖ(c)

= k ⋅ (c + ½(k+1))

= k ⋅ ½(2c + k + 1)

where ½(2c + k + 1) is integral.

Case 2 - k is even:

Sₖ(c)

= ½k ⋅ (2c + k+1)

where ½k is integral.

@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/

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/

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.

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.

@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.

@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/

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/

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/

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/

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

1/

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/

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/

3/

@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

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.

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:

Your husband will die tonight

unless we remove him from this hospital immediately.

I'll help.

The nurse lost her job;

but my Dad lived another 45 years.

Mrs. Geerkens:

Your husband will die tonight

unless we remove him from this hospital immediately.

I'll help.

The nurse lost her job;

but my Dad lived another 45 years.