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?  

$8n^2$ 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 $2(n-0.5)^2+0.5$.
