Problems & Puzzles: Conjectures

Conjecture 34. n & n2, Triangular Numbers

While visiting the wonderful page by Shyam Sunder Gupta, about Triangular Numbers I got interested by his affirmation that 1 & 6 are the only known triangular numbers such that their respective squares are also triangular numbers and his challenge to find the third one.

In personal emails interchanged between Shyam and I, he wrote:

"I can conjecture that there will not be any other (than 1 & 6) triangular number whose square is also a triangular number"

Question: Can you find a proof of the Shyam's conjecture or a counterexample? 

Contributions came from Sudipta Das, Joseph L. Pe, Carlos Rivera, Felice Russo & Johann Wiesenbauer

Sudipta Das wrote:

If T(n) is a square triangular number , then m( m + 1 ) / 2 = n2

Multiplying both sides of the equation by 8 gives ( 2m + 1 )2 - 2 * ( 2n )2 = 1

The above equation can be written in the form x2 - 2y2 = 1, where x is odd and y is even .

The smallest non-trivial solution is x = 3, y = 2 .                                                                                 

As x is odd and y is even , the recursive version is:

xi+1 = 3xi + 4yi
yi+1 = 3yi + 2xi

or the non-recursive:

xi = [ ( 3 + 2Ö2 )i + ( 3 - 2Ö2 )i ] / 2
yi = [ ( 3 + 2Ö2 )i  - ( 3 - 2Ö2 )i ] / ( 2Ö2 )

Note that (from the recursive solution) if xi is odd and yi is even, then xi+1 is odd and yi+1 is even , so that all the solutions (starting from the initial solution x = 3, y = 2) satisfy “x-odd ,y-even” condition .

Replacing xi by 2mi + 1 and yi by 2ni , we get : 

Initial values ( x = 3, y = 2 ) : m1 = 1, n1 = 1
Recursive formula :

mi+1 = 3mi + 4ni + 1
ni+1 = 3ni + 2mi + 1

or, non-recursively

mi = [ ( 3 + 2Ö2 )i + (3 - 2Ö2 )i - 2 ] / 4
ni = [ ( 3 + 2Ö2 )i - (3 - 2Ö2 )i ] / ( 4Ö2 )

The recursive formula is better for calculations as it involves integer calculations .

For ni to be a triangular number , 8ni + 1 must be a perfect square, i.e. Ö2 [ ( 3 + 2Ö2 )i - (3 - 2Ö2 )i ] + 1 must be a perfect square for some positive integer i .

The only values of i ( tested upto i = 3348 ) , for which 8ni + 1 is a perfect square ( or , ni is a triangular number ) are i = 1 ( ni = 1 ) and 2 ( ni = 6 ) .

Joseph L. Pe wrote:

Here's my proof of Shyam's conjecture, that 1 and 6 are the only triangular numbers which have squares that are also triangular...

First, I should acknowledge that the complete solution of the key Diophantine equation (*) below was shown to me by Ben Zimmer (bgzimmer@midway.uchicago.edu). It's an application of elliptic curve theory, and can be found at the Math Atlas site by Dave Rusin: http://www.math.niu.edu/~rusin/known-math/98/EC_integral... The solution is rather involved, but the main idea is to break down (*) into equations of lesser degree and solve their parametrizations.

Ok, let's have the proof.

A necessary and sufficient condition (NSC) for a number t to be triangular is: n(n+1)/2 = t for some positive integer n. Solving for n yields n = [sqrt(1+8t) - 1]/2 (The negative solution has been discarded.) Hence, an NSC for t to be triangular is that [sqrt(1+8t) - 1]/2 be an integer; equivalently, that sqrt(1+8t) be an odd positive integer.

Finding a triangular number which has a triangular square is equivalent to finding a positive integer m such that T(m)^2 is triangular, where T(m) = the m-th triangular number. Equivalently, sqrt(1 + 8 (m(m+1)/2)^2) is an odd integer; simplifying, sqrt(2m^4 + 4m^3 + 2m^2 + 1) is required to be an odd positive integer. Hence, an NSC for T(m) = m(m+1)/2 to have a triangular square is for the equation

(*) z^2 = 2m^4 + 4m^3 + 2m^2 + 1

to have an odd positive integer solution z.

Dave Rusin (see the above URL) has proved that the only positive integer solutions (m, z) of (*) are (m = 1, z = 3) and (m = 3, z = 17). (There are a few other integer--not necessarily positive integer--solutions.) These two values of m correspond to the triangular numbers T(1) = 1 and T(3) = 6

***

Carlos Rivera wrote:

It's well known (*) that all the triangular and square numbers are produced by the squares of the terms produced by the following recursive equation:

ui = 6*ui-1 -ui-2  where u1 =0 & u2=1

The sequence obtained is: 0, 1, 6, 35, 204, 1189, ...

So, the question involved in this conjecture is reduced to ask if is there a triangular number in this sequence, other than the first three (0, 1 & 6).

Computational verification can be easily done up to very large numbers ui, (**) but this (empirical) approach if far from demonstrating nothing, except than making the conjecture more plausible.

_____
(*) See p.197, Recreations in Theory of Numbers, A. H. Beiler

(**) With the help of the code sqrt.ub we have made search up to the limit of the Ubasic capabilities and the result is negative: no triangular number up to u2657 ( where u2657 has 2033 digits)

***

Felice Russo wrote:

Let's be x and x^2 two triangular numbers. This means that must exist two integers n and m such that: n(n+1)/2=[m(m+1)/2]^1/2 that implies: 2m^2+2m-[n(n+1)]^2=0 -->> m=[-1+/-2*sqrt(1+2(n(n+1))^2)]/2

Excluding the negative value we have: m=[-1+2sqrt(1+2*(n(n+1))^2)]/2

So in order to have x and x^2 triangular numbers must exist an integer n such that 1+2*(n(n+1))^2 is a perfect square. Equivalently the Shyam conjecture can be written as:

The only integers n such that 1+2*(n(n+1))^2 is a perfect square are 1 and 3.

Could be this second conjecture more easy to attach? With an Ubasic code I verified the conjecture up to n=377133374 that is for all triangular numbers up to 71114791080878625.

***

Johann Wiesenbauer wrote:

As for Shyam's conjecture concerning triangular numbers, I can provide some further evidence that it should be true though no proof. To be more precise, using the small Derive program below I have shown that any triangular numbers t>6 such that t^2 is also a triangular number must have more than 100,000 (!) digits.

test(s, u_ := [48, 8], p_) :=
LOOP(
IF(FIRST(u_) > 8s, exit),
u_ := [[6, -1]u_, FIRST(u_)],
p_ := 3,
LOOP(
IF(JACOBI(FIRST(u_) + 1, p_) = -1, exit),
p_ := NEXT_PRIME(p_),
IF(p_ > 101, RETURN FIRST(u_) + 1)))
 

test(10^100000) = true (651.8s)

Therein I used the following well-known facts about triangular numbers:

1. All triangular numbers that are also squares are given by the sequence t(1)=1, t(2)=6, t(n)=6t(n-1)-t(n-2) for n>2

(cf. http://www.madras.fife.sch.uk/maths/amazingnofacts/fact017.html )

2. An arbitrary natural number t is a triangular number n(n+1)/2 for some n if and only if 8t+1 is an odd square.

In my program above I first generated the slightly modified sequence u(n):=8t(n),n=1,2,... in the outer loop and checked in the inner loop the "square status" of u(n)+1 only mod p for all primes p <= 101. Since none of the numbers u(n)+1 passed this "probabilistic square test" for the given range 6<t(n)<10^100000, no further checking was necessary. (By the way, the upper bound 101 for the p's is minimal in this respect. When choosing 97 instead of 101 in a former run much to my surprise a huge "probable square" with 20902 digits showed up, which was "killed" though (sigh!) in a subsequent test using the very next prime p=101.)

***

Akash Betrieve wrote on March 12, 2020

I have listed two ways for you to obtain the solution to Conjecture no.34 and a way to obtain the solution to Puzzle 586.

1.   Check Shyam Gupta’s website: On his website, an acquaintance of his has found a solution to it via Mordell’s book on Diophantine Equations. http://www.shyamsundergupta.com/triangle.htm

2.    Or, you may also check the Fibonacci Quarterly where a paper by Luo Ming affirmatively resolves the problem in a very trivial fashion. Here is the link(No paywall): https://www.fq.math.ca/34-3.html  (Ming Luo

On the Diophantine Equation (x(x-1)/2)2 = (y(y-1)/2)2). I was pointed to this paper via Richard K. Guy’s book on Unsolved Problems in Number Theory 3rd edition 2004.- 

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