In how many ways 1387 can written in the sum of $n,(n>2)$ Consecutive natural numbers

In how many ways $1387$ can written in the sum of $n(n>2)$ Consecutive natural numbers?

1.$2$

2.$3$

3.$4$

4.$5$

First we can see that it can be written in the form of the sum of two Consecutive natural numbers.Try other cases.The answer is going to be $3$ but how can we prove it?If you find all three cases how can we be sure that there isn't any other?

Answer

HINT:

If $a$ is the first term of the $n$ consecutive natural numbers,
we have $$\dfrac n2\{2a+(n-1)\}=1387$$

$$\iff n^2+(2a-1)n-2\cdot1387=0$$

As $n$ is a natural number, the discriminant $(2a-1)^2+8\cdot1387$ has to be perfect square

Let $(2a-1)^2+8\cdot1387=(2b+1)^2$ where integer $a\ge1,2b+1>\sqrt{8\cdot1387}$

$\iff(b+a)(b-a+1)=2\cdot1387=2\cdot19\cdot73$

So, $b+a$ must divide $2\cdot19\cdot73$