The sum of the first $n$ $(n>1)$ terms of the A.P. is $153$ and the common difference is $2$. If the first term is an integer , then number of possible values of $n$ is
$a)$ $3$
$b)$ $4$
$c)$ $5$
$d)$ $6$
My approach : I used the formula for the first $n$ terms of an A.P. to arrive at the following quadratic equation
$n^2 + n(a-1) -153 = 0 $
Next up I realised that since we are talking about the number of terms , thus the
possible values which n can take must be whole numbers. That is the discriminant of the above quadratic should yield a whole number in other words
$ (a-1)^2 + 612 = y^2 $ for some y .
However I am stuck at this point , as from here I am unable to figure out the number of such a's ( i.e. the initial terms of an AP ) which will complete the required pythagorean triplet
The answer mentioned is $5$
Please let me know , if I am doing a step wrong somewhere . Or If you have a better solution , that will be welcomed too.
Best Answer
To summarize the (extensive) discussion in the comments:
The OP's method is sound and nearly complete. To finish it off we look at the relation $$612=y^2-(a-1)^2=(y+(a-1))(y-(a+1)$$ To solve that (over the integers) we simply need to factor $612=cd$ where the factors must have the same parity. There are three possible such factorings: $$\{18,34\},\;\{2,306\},\:\{6,102\}$$
Each of these gives rise to two possible starting points for our progressions. We get $$a\in \{-151,-47,-7,9,49,153\}$$
We reject the "degenerate" case $a=153$ as that progression just has a single term (and the OP specified $n>1$). Thus we have $5$ solutions.