<s>[INST] How many Integer values of $n$ are possible for $n^2+25n+19$ to be a perfect square. [1] How many Integer values of $n$ are possible for $n^2+25n+19$ to be a perfect square.
[2] How many Integer values of $n$ are possible for $n^2-19n+99$ to be a perfect square.
$\underline{\bf{My\;Try}}::$ for first one , Let $k^2 = n^2+25n+99$, where $k,n\in \mathbb{Z}$
So $4k^2 = 4n^2+100n+76\Rightarrow (2k)^2 = (2n)^2+2\cdot (2n)\cdot 25+625+(76-625)$
$(2k)^2 = (2n+25)^2-549\Rightarrow (2n+25)^2-(2k)^2 = 549 = 3^3\cdot 61$
Now Let $x= 2k$ and $y = (2n+25)$,
we get $(x^2-y^2)=(x+y)\cdot(x-y) = 3^2 \cdot 61$
Is it Right or not ,and is there is any other method to solve these type of questions,
If Yes the please explain here
Thanks
 [/INST] Let $$n^2+25n+19=(n+a)^2\text{ where }a\text{ is some integer }$$ 
$$\iff n=\frac{a^2-19}{25-2a}$$
Let integer $d$ divides both $a^2-19,25-2a$
$\implies d$ divides $\{2(a^2-19)+a(25-2a)\}=25a-38$
$\implies d$ divides $\{2(25a-38)+25(25-2a)\}=549=9\cdot61$
So, the divisors$(d)$ of $549$ are $\pm1,\pm3,\pm9\pm61,\pm183,\pm549$
As $25-2a$ must divide $a^2-19,$ check which values of $d(=25-2a)$ make $n$ integer
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