"If n is a triangular number, show that each of the three consecutive integers, $8n^2, 8n^2+1, 8n^2+2$ can be written as a sum of two squares."
I have spend hours working on this problem and cannot seem to get anywhere with it. I was advised to start with $8n^2+1$ and work through it just using algebra to express this as a sum of two squares, but I am really struggling. Can anyone help?
On
HINT:
If $n$ is a triangle number, than there is a $k$ that gives you $n = \frac{k*(k+1)}{2},$ which then gives us
$8n^2 = 2*k^2*(k+1)^2 = a^2+b^2$
$8n^2$ is even, so a and b are either both even or both odd.
$8n^2+1 = 2*k^2*(k+1)^2+1 = c^2+d^2 = a^2+b^2+1$
$8n^2 + 1$ is odd, so "c" is even and "d" is odd.
$8n^2+2 = 2*k^2*(k+1)^2 = e^2+f^2 = c^2+d^2+1 = a^2+b^2+2$
$8n^2+2$ is even, so "e" and "f" are either both even or both odd.
Does this give you an idea for what to do next?
On
It is assumed in the books on this subject that all square $a^2$ is a sum of two squares $a^2+0^2$.
► Note first that for all $n$ one has $$8n^2=(2n)^2+(2n)^2\qquad (1)$$ Now all triangular number $n$ is of the form $$n=\frac{m(m+1)}{2}$$ Hence $$8n^2=\frac{8m^2(m+1)^2}{4}=2(m^2+m)^2$$ It follows the two identities
$$2(m^2+m)^2+1=(m^2-1)^2+(m^2+2m)^2\qquad (2)$$ $$2(m^2+m)^2+2= (m^2+m+1)^2+(m^2+m-1)^2\qquad (3)$$
With $(1),(2),(3)$ we have verified that the statement is true.
$8n^2$ is trivial, since $8n^2=(2n)^2+(2n)^2$.
For the first triangular numbers, we have
Note that the numbers $0,3,8,15,24,\ldots$ form the sequence $\{n^2-1\}$.
Do something similar for $8n^2+2$.
EDIT (answering an OP's comment):
The "small" squares sequence is $1,5,11,19,\ldots$, which is $2(n-0.5)^2+0.5$.