11 Mar, 10 tweets, 5 min read
@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).
@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.
@GWOMaths In both cases the value of Sₖ(c) always has an odd factor greater than 1, either k itself in Case 1 or (2c + k+1) in Case 2.

Therefore all powers of 2 are inexpressible as any such Sₖ(c).

Likewise all odd primes are inexpressible as c would be negative.
@GWOMaths There are 11 primes and powers of 2 between 50 and 100 inclusive; namely:
53, 59, 61, 64, 67, 71, 73, 79, 83, 89, 97.

It remains to show that all remaining values of n in the range ARE expressible as such an Sₖ(c).
@GWOMaths For any such n ∈ Z⁺ with 50 ≤ n ≤100:
let p be its smallest odd prime factor (and clearly ≥ 3).

Case: n/p ≥ ½(p+1)

Then in order
2⋅(n/p) ≥ p+1;

2⋅(n/p) - (p+1) ≥ 0;

and since
2⋅(n/p) - (p+1)
is even and positive

then
n = Sₚ( ½(2⋅(n/p) - (p+1)) )
satisfies.
@GWOMaths Case: n/p < ½(p+1)

Note:
If n has any other prime factor, whether p with multiplicity greater than 1 or a q > p, then

n/p ≥ p ≥ ½(p+1)
and we are in the case above.

So this case only occurs when
n = p⋅2ᵐ
for some m > 0.

Then
n/p = 2ᵐ < ½(p+1)

and
2ᵐ⁺¹ < p+1 ≤ p-1.
@GWOMaths So
p - 1 - 2ᵐ⁺¹ ≥ 0

and, since p is odd,
½(p - 1 - 2ᵐ⁺¹)

is integral and positive or zero.

Setting k = 2ᵐ⁺¹ = 2⋅(n/p)
Sₖ(p - 1 - k)
= ½k ⋅ ((p-1-k) + k+1)
= 2ᵐ ⋅ (p)
= n.

Q.E.D.

Note that all positive integers < 6 are either 1; a prime; or a power of 2.
@GWOMaths Therefore all numbers n with 50 ≤ n ≤ 100 are expressible as a sum of at least 3 consecutive integers except the 11 numbers which are prime or powers of 2.

Not as elegant as the answer I look forward to seeing tomorrow - but it got the job done.

😀

• • •

Missing some Tweet in this thread? You can try to force a refresh

This Thread may be Removed Anytime!

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

# More from @pgeerkens

8 Mar
@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.
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.