<s>[INST] $A+B+C=2149$, Find $A$ In the following form of odd numbers 

If the numbers 
taken from the form where $A+B+C=2149$
Find $A$
any help will be appreciate it, thanks.
 [/INST] Looking at the first element in each row, we see
 row      A    B    C   A+B+C
   1      1    3    5       9
   2      3    7    9      19
   3      7   13   15      35
   4     13   21   23      57
   5     21   31   33      85
   6     31   43   45     139

Using finite differences, we find that 
\begin{align}
   A &= n^2 - n + 1\\
   B &= n^2 + n + 1 = A+2n\\
   C &= n^2 + n + 3 = A+2n+2\\
A+B+C &= 3n^2+n+5
\end{align}
where A is the first element in row $n$.
For each shift of A to the right in the same row, A,B, and C increase by 2
and A+B+C increases by $6$.
Solving $3n^2+n+5 = 2149$ for n, we get $n = 26.567$ So we are in row $26$.
For the first element in row $26,
 A = 651,\; B = 703,\; C = 705,$ and $A+B+C = 2059$.
Since $\dfrac{2149-2059}6 = 15$, we need to increase $A, B, $ and $C$ by $2\cdot15 = 30$.
So
\begin{align}
   A &= 681\\
   B &= 733\\
   C &= 735
\end{align}
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