How can I find integer values for which a given expression gives a perfect square?

Question

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)?

Answer 1

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

Answer 2

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 $2x+19 = \pm 3$.

Answer 3

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.

Answer 4

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 $.