Triangular numbers for numbers.

Question

Interestingly for triangular numbers:

$X(X+1)+Y(Y+1)=Z(Z+1)+a$

$a$ - this number is determined by the condition of the problem. Are all numbers equation has a solution? And what kind of formula in general is?

Answer 1

Gauss proved that every number is a sum of three triangular numbers. I don't think there's a good formula for the summands.

Answer 2

Well why not?

If the equation: $X(X\pm1)+Y(Y\pm1)=Z(Z\pm1)+a$

The number can be set as follows: $a=\frac{q^2-1}{4}$

Then using the appropriate equation Pell: $p^2-2s^2=\pm1$

Solutions can be written as:

$X=(2L+1+q)s^2+(q+2L)ps-(\frac{1-q}{2})p^2$

$Y=(1+q)s^2+(q+2L)ps+(L-\frac{1-q}{2})p^2$

$Z=(2L+1+q)s^2+2(q+L)ps+(L-\frac{1-q}{2})p^2$

Or.

$X=(2L+1-q)s^2+(q-2L)ps-(\frac{1+q}{2})p^2$

$Y=(1-q)s^2+(q-2L)ps+(L-\frac{1+q}{2})p^2$

$Z=(2L+1-q)s^2+2(q-L)ps+(L-\frac{1+q}{2})p^2$

$L$ - any integer given by us. And all the numbers can be any character.

Answer 3

If the equation:

$X(X+1)+Y(Y+1)=Z(Z+1)+q$

We write the solution if the number of: $q=(a-b)(a+b+1)$ This can be represented as ideal:

Then use the equation Pell: $p^2-2k(k-t)s^2=1$

$k,t$ - integers asked us.

Then the solution can be written:

$X=((2b+1)k-(2a+1)t)ps+2((a+1)k^2-(a+b+1)kt+bt^2)s^2$

$Y=ap^2+((2b+1)k-2bt)ps+2(k^2+(b-a-1)kt)s^2$

$Z=bp^2+(2(a+1)k-(2a+1)t)ps+2((b+1)k^2-(a+b+1)kt+bt^2)s^2$

Or again:

$X=(2b-2a-1)p^2+((6b-4a-1)k+(2a-4b+1)t)ps+$$2((2b-a)k^2-(3b-a)kt+bt^2)s^2$

$Y=(2b-a-1)p^2+((6b-4a-1)k-2bt)ps+2(2(b-a)k^2-(b-a)kt)s^2$

$Z=(3b-2a-1)p^2+((8b-6a-2)k+(2a-4b+1)t)ps+$ $+2((3b-2a)k^2-(3b-a)kt+bt^2)s^2$