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First things first. I've made this a seperate thread off my previous topic so that these topics don't conflict if I'd posted this to my previous question.

I believe any $k$ triangular numbers will have a unique sum, such that they can not be composed by any other $k$ triangular numbers. I believe my conjecture is right, here's my proof (other than experimenting);

Take two triangular numbers $(k=2)$ $a, b$ and $c, d$. Assuming the conjecture is false, the equation must prove;

$a(a+1)/2 + b(b+1)/2=c(c+1)/2 + d(d+1)/2$

which implies (after simplification) $a + b + a^2 + b^2 = c + d + c^2 + d^2$

This means that the natural numbers $a, b$ and $c, d$ must be equal to each other's squares, as well as their sum I'm not at all sure about this part...

Maybe that isn't even a valid proof, but if someone thinks the conjecture is right or wrong, please bother to give a proof why!

Important extra information: Fine, I've got all answers for k=2 uptil now, what if k were to be 3, 4 or greater? Are there possibilities then?
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$3(3+1)/2 + 4(4+1)/2 = 6 + 10 = 16 = 1 + 15 = 1(1+1)/2 + 5(5+1)/2$. – Rahul Apr 4 '12 at 11:48
Oh right. I'd overlooked that completely. – Mach9 Apr 4 '12 at 11:52
Is that the only one there? I've just about finished experimenting upto the 19th triangular numbers (well, I don't deny having missing something...) – Mach9 Apr 4 '12 at 11:57
up vote 2 down vote accepted

If you allow 0 as a triangular number then there is plenty of examples showing that there are numbers which have more than one expression as sum of two triangular numbers.


There are infinitely many solutions of $\triangle_a=2\triangle_b$, which is the same as $\triangle_0+\triangle_2=\triangle_b+\triangle_b$. This equation is closely related to the problem of finding square triangular numbers, i.e. solutions of $\triangle_c=d^2$. See this question.

Another infinite class of solutions with one of the triangular numbers being 0 was given in an answer to this question.

But there are also examples where some of the triangular numbers is non-zero, some of them can by find by inspection checking the sequence of triangular numbers, see A000217 at OEIS.



Although your conjecture is not true, the question of characterizing $a$, $b$, $c$, $d$ with $\triangle_a+\triangle_b=\triangle_c+\triangle_d$ seems to be interesting anyway.

Note also that your equation $a^2+a+b^2+b=c^2+c+d^2+d$ is equivalent to
So this is the same thing as asking for the numbers that have more than one expression as sum of two squares of odd numbers.

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Not exactly triangular numbers, but just about achieves my purpose... Here it is. Can someone modify this to represent triangular numbers? – Mach9 Apr 4 '12 at 12:05
What I mean from this is, will any $k$ numbers in the sequence be exclusively the sum of only those $k$ numbers? And can I simply apply $T(n$th element of sequence in link$)$ to make all numbers triangular? – Mach9 Apr 4 '12 at 12:08

Since every positive integer is the difference between two consecutive triangular numbers, you can find plenty counterexamples. For instance $\binom72-\binom52=21-10=11=\binom{12}2-\binom{11}2$ so $$ \binom72+\binom{11}2=76=\binom{12}2+\binom52 $$

Added: more generally for any $k\geq2$, write down two random sums of $k-1$ triangular numbers, and compute their difference $d$ (taken in the direction such that $d\geq0$). Now add a final trangular number $\binom d2$ to the larger sum and $\binom{d+1}2$ to the smaller sum, to obtain equality.

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What if $k$ were to be more than 2? – Mach9 Apr 5 '12 at 6:45
See the added part. – Marc van Leeuwen Apr 5 '12 at 13:06

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