Triangular Numbers and Sum of Two Squares "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?  
 A: 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.
A: $8n^2$ is trivial, since $8n^2=(2n)^2+(2n)^2$.
For the first triangular numbers, we have


*

*$8\cdot1^2+1=9=0^2+3^2$

*$8\cdot3^2+1=73=3^2+8^2$

*$8\cdot6^2+1=289=8^2+15^2$

*$8\cdot10^2+1=801=15^2+24^2$

*...


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):


*

*$10=1^2+3^2$

*$74=5^2+7^2$

*$290=11^2+13^2$

*$802=19^2+21^2$

*...


The "small" squares sequence is $1,5,11,19,\ldots$, which is $2(n-0.5)^2+0.5$.
A: Note that 
$$8n^2=(2n)^2+(2n^2)$$
and 
$$8n^2+2=(2n-1)^2+(2n+1)^2$$
so it is both $8n^2$ and $8n^2+2$ can be expressed as the sum of two squares, whether or not $n$ is a triangular number.
A: 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?
