Unique sum of $k$ triangular numbers into $k$ triangular numbers?

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

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.

$0+21=6+15$
$\frac{0\cdot1}2+\frac{6\cdot7}2=\frac{3\cdot4}2+\frac{5\cdot6}2$

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.

$1+15=6+10$
$\frac{1\cdot2}2+\frac{5\cdot6}2=\frac{3\cdot4}2+\frac{4\cdot5}2$

$10+21=3+28$
$\frac{4\cdot5}2+\frac{6\cdot7}2=\frac{2\cdot3}2+\frac{7\cdot8}2$

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