2024/031) Find triangular number which is one less than a multiple of 11.

 We have $n^{th}$ triangular number  $t_n= \sum_{k=1}^n k = \frac{n(n+1)}{2}$

 So we must have  $\frac{n(n+1)}{2} \equiv -1 \pmod {11}$

Or $n(n +1) \equiv -2 \pmod {11}$

Or $n^2 + n + 2 \equiv 0 \pmod {11}$

Or $4 n^2 + 4n + 8 \equiv 0 \pmod {11}$ (the purpose of doing this is to covert to perfect square as evident from next line)

Or $(2n+1)^2 + 7 \equiv 0 \pmod {11}$

Or  $(2n+1)^2 =  \equiv -7 \pmod {11}$

Or  $(2n+1)^2 =  \equiv 4 \pmod {11}$

Let us find the square mod 11 for n = 0 to 5 we get (0,0),(1,1),(2,4),(3,9),(4,5),(5,3)$

So the numbers are 2 and 9 ( that is 11 -2)

$2n + 1 \equiv  2  \pmod {11}$ gives n = 6 and $2n + 1 \equiv  9  \pmod {11}$ gives n = 4

So we have the triangular numbers are $t_{11k+4}$ and $t_{11k+6}$ for any non negative k