Calculation of Integral values of $x$ for which the expression $x^2+19x+92$ is a perfect square.
$\bf{My\; Solution::}$ Let $x^2+19x+92 = k^2\;,$ where $x,k\in \mathbb{Z}$
$$x^2+19x+(92-k^2)=0$$
Now $$\displaystyle x = \frac{-19\pm \sqrt{19^2-4(92-k^2)}}{2} = \frac{-19\pm \sqrt{4k^2-7}}{2}$$
Now for perfect square $\displaystyle 4k^2-7 = 0\Rightarrow k=\pm \frac{\sqrt{7}}{2}$
But it is wrong,
Please explain me how can i solve it and where i am wrong,
thanks
Easier than directly using the quadratic formula: $$\eqalign{ y^2=x^2+19x+92\quad &\Leftrightarrow\quad 4y^2=4x^2+38x+368=(2x+19)^2+7\cr &\Leftrightarrow\quad 4y^2-(2x+19)^2=7\cr &\Leftrightarrow\quad (2y-2x-19)(2y+2x+19)=7\cr}$$ and so on.
You need: $4k^2 - 7 = n^2 \to 4k^2 - n^2 = (2k-n)(2k+n) = 7$. This gives:
$2k - n = 1$, and $2k + n = 7$. So: $4k = 8$, and $k = 2$, and $n = 3$. Thus: $x = \dfrac{-19 \pm 3}{2} = -11$ or $-8$.
$2k - n = - 7$, and $2k + n = -1$. Thus: $4k = -8$, and $k = -2$. Thus: $n = 3$, and we have: $x = - 11$ or $-8$.
$2k + n = -7$, and $2k - n = -1$. Then: $4k = -8$, and $k = -2$. Thus: $n = -3$. And this also gives the same values of $x$ and $k$ as above.
The same is true for the case: $2k - n = 7$,and $2k + n = 1$
Firstly, I would "complete the square":
$$\left(x + \frac{19}{2}\right)^2 + 92 - \left(\frac{19}{2}\right)^2 = k^2$$
Multiplying throughout by $4$, this gives us
$$(2x + 19)^2 + 7 = (2k)^2$$
Rearranging and doing some factorization,
$$(2k + 2x + 19)(2k - 2x - 19) = 7$$
To carry on, notice that $7 = 1 \cdot 7 = -1 \cdot -7$. Hence the possible candidates are
$$2k + 2x + 19 = \pm 1, 2k - 2x - 19 = \pm 7$$
or
$$2k + 2x + 19 = \pm 7, 2k - 2x - 19 = \pm 1$$
Simply solve each of the four cases to determine the correct value of $x$.
Write: $$x^2+19x+92 =y^2$$
If $x>-8$ we have: $$x^2+19x+92 < x^2+20x+100 = (x+10)^2$$
and if $x>-11$ we have: $$(x+9)^2 = x^2+18x+81<x^2+19x+92$$
So for $x>-8$ we have $$(x+9)^2<x^2+19x+92<(x+10)^2$$ or after taking square root $$|x+9|<|y|<|x+10|$$ and thus we have no integer solution for $x>-8$.
Say $x<-7$. Then we can write $x=-t$ where integer $t>7$. So we have $$t^2-19t+92 =y^2$$ and we procedee simmilary...
Let $\displaystyle x^2+19x+92=(x+a)^2\iff x=\frac{a^2-92}{2a-19}$
If integer $d$ divides both $a^2-92,2a-19$
$d$ must divide $2(a^2-92)-a(2a-19)=19a-184$
$d$ must divide $19(2a-19)-2(19a-184)=7$
The necessary condition for $x$ to be integer is $2a-19$ must divide $7$
Check for the values of $2a-19$
