$1,3,6,10,...,2016$ IE THE TRIANGULAR NUMBERS

$8n+1 = a^2$ for an integer a Thus, $n=\frac{a^2-1}{8}$. Hence, $a^2-1 \equiv 0 (\mod 8)$ $a^2 \equiv1 (\mod 8)$ $a=1, 3, 5, 7, \dots$. We see then that $n$ corresponds to the values of $a=1, 2, 3 \dots$. The first few are: $n = 0, 1, 3, 6...$. Indeed, the triangular numbers.

Note that we wish to seek the integer $n$ such that $n=1\mod{8}$. Observe that $8n+1=m^2$ is what we seek such that $m$ is some number. Using what we have, note that $(2k+1)^2\equiv 1 \mod{8}$. Thus, each of the odd squares up to $43^2$ can be written as $8n+1$. $43^2=1849$. Hence, there are $21$ such $n$.

^ above answer very sus

imagine calling an answer sus when yours is sus :flooshed:

@grolarbear Your statement that $a^2-1 \equiv 0 \pmod 8$ means $a^2 \equiv 1 \pmod 8$ is incorrect. Any odd number satisfies $a^2-1 \equiv 0 \pmod 8$.

OlympusHero wrote: @grolarbear Your statement that $a^2-1 \equiv 0 \pmod 8$ means $a^2 \equiv 1 \pmod 8$ is incorrect. Any odd number satisfies $a^2-1 \equiv 0 \pmod 8$. Erm, exactly why is it incorrect? @grolarbear simply add 1 to both sides... @below haha well, just be careful

Sorry, I forgot how to read and thought he said $a \equiv 1 \pmod 8$. Apparently skipping first grade is starting to take its toll six grades later.