Proof? hex, square, triangular numbers

[Frank Knorr]

Is there a proof that no Centered Hexagonal Numbers (greater than one) can also be both a Square Number and a Triangular Number? 

For positive integers a, b, & c: 

Centered Hexagonal Number (aka hex number) = 1 + 3 x a x ( a - 1 ) = a^3 - ( a - 1 )^3 

and also = 1 + 6 x Triangular Number = 1 + 6 x ( a - 1 ) x ( ( a - 1 ) + 1 ) / 2;

Square Number = b x b; and

Triangular Number = c x ( c + 1 ) / 2.

In https://mathworld.wolfram.com/HexNumber.html I found the statement:

“The only hex number that is both square and triangular is 1”.

And, on page 32 of Leonard Eugene Dickson’s History of the Theory Of Numbers, Volume II, Diophantine Analysis (citing an 1898 article):

“J. W. West noted that if [a triangular number] = 6 x [a different triangular number] + 1 

[that first triangular number] is not a square”.

Since “6 x [a triangular number] + 1” is a hex number these two quotes are equivalent , but neither quote is accompanied by a proof.  I have tried unsuccessfully to find such a proof and would appreciate even a hint about how to approach this problem.

I am a fairly unsophisticated (aging) amateur mathematician. From time to time I have dabbled with relationships among triangular, square, and hex numbers.

Triangular Squares (https://oeis.org/A001110)

I found that, after the trivial case of 1, the next 

triangular number that is also a square number is 36. 

The 8th triangular number (8x9/2) is equal to the 6th square number (6x6). 

The notation I have adopted is: 

TS(1)=1, TS(2)=36, ts(1)=1, ts(2)=8, st(1)=1, and st(2)=6. 

Further triangular numbers that are also square numbers can be generated(*) using:

TS(n) = 34 x TS(n-1) - TS(n-2) + 2

ts(n) = 6 x ts(n-1) - ts(n-2) + 2

st(n) = 6 x st(n-1) - st(n-2)

Also, a better and better approximation of the square root of 2 can be obtained by dividing ts(n) by st(n) for larger & larger values of n. In fact, 

ts(n) = sqrt(2) x st(n) - 0.5 + e, 

where e is “a bit more” which gets to be very small as n increases (e.g. e is less than 0.000013 when n = 6). 

Both ratios, 

ts(n+1)/ts(n) and st(n+1)/st(n), approach 3 + 2 x sqrt(2) as n increases, 

while TS(n+1)/TS(n) approaches 17 + 12 x sqrt(2).

By setting {xts}(n) = 2 x ts(n) + 1 and {yst}(n) = 2 x st(n), the pell equation

{xts}(n)^2 - 2 x {yst}(n)^2 = 1, with {xts}(1) = 3 and {yst}(1) = 2

can be used to generate additional values of {xts}(n) and {yst}(n). 

However, I am unable to prove that TS(n) cannot also be a hex number for n>1.

Triangular Hexagons (https://oeis.org/A006244)

Similarly, after the trivial case of 1, the next 

triangular number that is also a hex number is 91. 

The 13th triangular number (13x14/2) is equal to the 6th hex number (1+3x6x5). 

Using my notation TH(1)=1, TH(2)=91, th(1)=1, th(2)=13, ht(1)=1, and ht(2)=6. 

Triangular numbers that are also hex numbers can be generated(*) using:

TH(n) = 98 x TH(n-1) - TH(n-2) - 6 

th(n) = 10 x th(n-1) - th(n-2) + 4

ht(n) = 10 x ht(n-1) - ht(n-2) - 4

Also, a better and better approximation of the square root of 6 can be obtained by dividing th(n) by ht(n) for larger & larger values of n. In fact, 

th(n) = sqrt(6) x ht(n) - 0.5 x sqrt(6) - 0.5 + e, 

where e is a small amount that gets to be very small as n increases (e.g. e is less than 0.0000029 when n = 6). 

Both ratios, 

th(n+1)/th(n) and ht(n+1)/ht(n), approach 5 + 2 x sqrt(6) as n increases, 

while TH(n+1)/TH(n) approaches 49 + 20 x sqrt(6).

By setting {xth}(n) = 2 x th(n) +1 and {yht}(n) = 2 x ht(n) - 1, the equation

{xth}(n)^2 - 6 x {yht}(n)^2 = 3 and the pell equation 

{xth}(n)^2 - 6 x {yht}(n)^2 = 1, 

with {xth}(1) = 3 and {yht}(1) = 1 and {xth}(2) = 27 and {yht}(2) = 11

can be used to generate additional values of {xth}(n) and {yht}(n). 

However, I am unable to prove that TH(n) cannot also be a square number for n>1.

Square Hexagons (https://oeis.org/A006051)

I also found that, after the trivial case of 1, the next 

square number that is also a hex number is 169. 

The 13th square number (13x13) is equal to the 8th hex number (1+3x8x7). 

With my notation SH(1)=1, SH(2)=169, sh(1)=1, sh(2)=13, hs(1)=1, and hs(2)=8. 

Additional square numbers that are also hex numbers can be generated(*) using:

SH(n) = 194 x SH(n-1) -SH(n-2) -24

sh(n) = 14 x sh(n-1) - sh(n-2)

hs(n) = 14 x hs(n-1) - hs(n-2) - 6 

A better and better approximation of the square root of 3 can be obtained by dividing sh(n) by hs(n) for larger & larger values of n. In fact, 

sh(n) = sqrt(3) x hs(n) - 0.5 x sqrt(3) + e, 

where e is a tiny amount that gets smaller & smaller as n increases (e.g. e is less than 0.0000026 when n = 6). 

Both ratios, 

sh(n+1)/sh(n) and hs(n+1)/hs(n), approach 7 + 2 x sqrt(12) as n increases, 

while SH(n+1)/SH(n) approaches 97 + 28 x sqrt(12).

By setting {xsh}(n) = 2 x sh(n) and {yhs}(n) = 2 x hs(n) - 1, the pell equation

{xsh}(n)^2 - 3 x {yhs}(n)^2 = 1, with {xsh}(1) = 2 and {yhs}(1) = 1

can be used to generate additional values of {xsh}(n) and {yhs}(n). 

However, I am unable to prove that SH(n) cannot also be a triangular number for n>1.

[(*) using Math Induction, I have shown that these formulas generate all the desired values.]

Values with as many as 150 digits were examined but no number was found that is all three: hex, square, and triangular. 

In an attempt to find such a counter-example greater than 10^150 further tests were tried.

By setting

u = 5 + 2 x sqrt(6) and 

w = 7 + 2 x sqrt(12)

values of ht(n) and hs(n) can be estimated for ever greater values of n.

To minimize estimation errors of smaller numbers, the base of projecting such values could use fixed / initial values of i and j :

ht(f) = ht(i) x u^( f - i ); so, log( ht(f) ) = log( ht(i) ) + ( f - i ) x log( u ); 

hs(k) = hs(j) x w^( k - j ); and therefore, log( hs(k) ) = log( hs(j) ) + ( k - j ) x log( w ).

Then, values of ht(f) and hs(k) can be approximate and compared. If there is an ht(f) that is (almost) equal to an hs(k), that might indicate that a hex number is both triangular and square.

Simply by testing whether k is (very close to) an integer for various values of f, candidates might be identified:

k = ( f x log(u) - i x log(u) + j x log(w) + log( ht(i) ) - log( hs(j) ) / log(w)

For example, 

log(u) = 0.995590424230678

log(w) = 1.14389509506672

I = j = 10

log ( ht(i) ) = 8.70553404145927

log ( hs(j) ) = 10.02638278446647

produces values of k that are nearly integers when 

f = 55 ( k is close to 48 ), 

f = 433 ( k is close to 377 ),

f = 1428 ( k is close to 1243 ),

and many other values.  But the limitations of Excel make accurate calculations difficult.

How can I be sure that there is not some very large number that is all three?

Frank Knorr

141 Echo Drive

Vernon, CT 06066

mandma...@yahoo.com

fekn...@gmail.com

[Dave Consiglio]

A quick AI search (yeah, I know) gives this:

https://www.google.com/search?q=are+there+numbers+that+are+square%2C+triangular%2C+and+centered+hexagonal+at+the+same+time&sca_esv=f1e464e6e7e851fc&sxsrf=APpeQnuinDjU2yRE7ov0fNDyiMYo4LNV8w%3A1787101712756&ei=9gGFao2iAdyT0PEPpYetmAY&biw=1470&bih=808&uact=5&sclient=gws-wiz-serp&udm=50&fbs=ABfTbFVyMZGZf1hfvX9uKjN_-G8c4u0nXx4bEIpwm1lnNH832a9BVCEiB2iPJNekNderQwLP8msUKsz-6AMxGyueJZ9cK2mNBy2II_6WvkKSZWdegmv0MXIJU8fN2UGC6Y60IgSnyIHDybjdnHbnQzqoO6AovirrMTfe19eou43y2idgp4VPoywv5V__0waMiE1GMb4p6RS8lF2S7Si2dPtKxXzn_zfltg&aep=10&ntc=1&mstk=AUtExfC3D9xJrXnuM4wb0KaOvKGh_I7yVEuWhlQm5Ct_nkoQo741KKPH4obWe4LFb7ZumO-LUFNmF42LJeXogkUAQ2alDLZnni-3C-Oxi_YlK7oQwSJccVRTCrk2DmGagUk48WGTXFBnp4aXlm3q4tEU9bOEjN1MIvZCDcFd0LHLm6uuWVJIu7sKf9g6ViFrfpFH1A9DHdex17_hrEl7KlFuNQhEFqnITLT-ZjVWz7x1E0Sy9s9ADDGWdqsOzeNLsWoY2-u5CVeY91Nyr0w5rriXtDVLN6LmD_jHlFaakgXbYCOAn0hr_j_xl7-EmIkZf88_q0HF_O4pE1rVjA&aioh=3&csuir=1&cs=1&atvm=2&mtid=KgKFaqSdLKa_ruEPj6fh0Aw

Summarized:

In 1978, mathematician Charles Grinstead published a rigorous proof in Mathematics of Computation specifically solving this class of equations. Using advanced algebraic number theory (including linear forms in logarithms and Baker's method), Grinstead proved that the only non-negative integer solution to this system is:

(s=1, y=3, z=1)

I doubt I will ever understand this level of mathematics (or, for that matter, how AI really works), but it seems that Grinstead's work may be a good place to look.

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[D. S. McNeil]

There's also "ON A FAMILY OF QUARTIC EQUATIONS AND A DIOPHANTINE PROBLEM OF MARTIN GARDNER" by P.G. Walsh at https://web.math.pmf.unizg.hr/~glasnik/vol_41/no2_04.html, where Walsh uses Ljunggren's theorem instead.


Doug

[Frank Knorr]

Thanks Doug. I appreciate the quick response.

Frank

[Frank Knorr]

Thanks Dave. I will definitely look into these sources.

Frank