How can I find integer values for which a given expression gives a perfect square? Find the integer values for which $x^2+19x+92$ is a perfect square.
Also, How to proceed if you have to find values ( not necessarily integer)?
 A: Complete the square:
$$\begin{align}
x^2 + 19x + 92 &= m^2\\
\iff (2x)^2 + 2\cdot 19\cdot (2x) + 368 &= (2m)^2\\
\iff (2x+19)^2 + 7 &= (2m)^2\\
\iff (2m)^2 - (2x+19)^2 &= 7.
\end{align}$$
Since $7$ is prime, that means $2m = \pm 4$ and $2x+19 = \pm 3$.
A: I prefer to bound the expression between two consecutive squares. We can show that when $ x > - 8$,
$$ (x+9)^2 < x^2 + 19x + 92 < (x+10)^2. $$
When $x < - 11$.
$$ (x+9)^2 > x^2 + 19x + 92 > (x+10)^2. $$
Hence, we only need to test $x = -11, -10, -9, -8 $.
A: Let $x^2+19x+92=(x+a)^2$ where $a$ is any real number
$\displaystyle\implies x=\frac{a^2-92}{19-2a}$
If you allow non-integer values, for any real value of $a, x^2+19x+92$ will be perfect square
If we restrict the solution to integers only,   $a$ has to be integer (why?)
Let integer $d$ divides  both $a^2-92,19-2a$
$\displaystyle\implies d$ divides $2(a^2-92)+a(19-2a)=19a-184$
$\displaystyle\implies d$ divides $19(19-2a)+2(19a-184)=7$
$\implies (a^2-92,19-2a)|7$
So, the necessary condition of $x$ being integer is the denominator $19-2a$ must divide $7$
$\displaystyle\implies 19-2a=\{\pm1,\pm7\}$
Check for all the four cases
A: You could look at writing this as equal to $(x+9)^2 =x^2+18x+81$.  Put this equal to your expression, and you get $x+11=0$.  Similarly, other polynomials give similar results.
