Problems & Puzzles:
Conjectures
Conjecture
34.
n
& n^{2}, 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 = n^{2}
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 x^{2}  2y^{2} = 1, where x is odd
and y is even .
The smallest nontrivial
solution is x = 3, y = 2
.
As x is odd and y is even , the
recursive version is:
x_{i+1} = 3x_{i}
+ 4y_{i}
y_{i+1} = 3y_{i} + 2x_{i}
or the nonrecursive:
x_{i} = [ ( 3 + 2Ö2
)^{i} + ( 3  2Ö2
)^{i} ] / 2
y_{i} = [ ( 3 + 2Ö2
)^{i}  ( 3  2Ö2
)^{i} ] / ( 2Ö2
)
Note that (from the recursive
solution) if x_{i} is odd and y_{i} is even, then x_{i+1}
is odd and y_{i+1} is even , so that all the solutions (starting
from the initial solution x = 3, y = 2) satisfy “xodd ,yeven”
condition .
Replacing x_{i} by 2m_{i}
+ 1 and y_{i} by 2n_{i} , we get :
Initial values ( x = 3, y = 2 )
: m_{1} = 1, n_{1} = 1
Recursive formula :
m_{i+1} = 3m_{i}
+ 4n_{i} + 1
n_{i+1} = 3n_{i} + 2m_{i} + 1
or, nonrecursively
m_{i} = [ ( 3 + 2Ö2
)^{i} + (3  2Ö2
)^{i}  2 ] / 4
n_{i} = [ ( 3 + 2Ö2
)^{i}  (3  2Ö2
)^{i} ] / ( 4Ö2
)
The recursive formula is better
for calculations as it involves integer calculations .
For n_{i} to be a
triangular number , 8n_{i} + 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 8n_{i} + 1 is a perfect square ( or
, n_{i} is a triangular number ) are i = 1 ( n_{i }= 1 )
and 2 ( n_{i} = 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/knownmath/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 mth 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 integernot necessarily positive
integersolutions.) 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:
u_{i} = 6*u_{i1}
u_{i2
}where u_{1}
=0 & u_{2}=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
u_{i},
(**) 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 u_{2657}_{
}( where u_{2657}
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 wellknown 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(n1)t(n2) 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.
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/343.html (Ming
Luo
On the
Diophantine Equation (x(x1)/2)2 = (y(y1)/2)2). I was pointed to
this paper via Richard K. Guy’s book on Unsolved Problems in Number
Theory 3rd edition 2004.
***
