# positive integer value of $n$ for which $n^2-19n+99$ is a perfect square.

Calculate positive integer value of $n$ for which $n^2-19n+99$ is a perfect square.

My Try:: Let $n^2-19n+99 = k^2$ where $k\in \mathbb{Z}$

$4n^2-76n+396 = 4k^2$ or $(2n-19)^2-35 = (2k)^2$

$(2n-19)^2-(2k)^2 = 35$

now we have to take two perfect square whose difference is $= 35$

so one pair is $(6^2,1^2)$

now my question is how can i calculate for other ordered pairs

Thanks

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$(2n-19)^2-(2k)^2 = -35$

$\Rightarrow (2n-19-2k)(2n-19+2k)=35$

$35=5\times 7=35\times 1=-5\times -7=-35\times -1$

Check the possible pairs.

Another way: $n^2-19n+99 = k^2$ is a quadratic in n so for it to have integral(perhaps rational) roots its determinant is a square.

$\Rightarrow 19^2-4(99-k^2)=y^2,y\in \mathbb{N}$

$\Rightarrow19^2-4.99=y^2-4k^2\Rightarrow 361-396 \Rightarrow35=(2k)^2-y^2=(2k+y)(2k-y)$

$35=5\times 7=35\times 1=-5\times-7=-35\times -1$

$\Rightarrow 2k+y=5$

$\Rightarrow 2k-y=7$

$\Rightarrow 4k=12\Rightarrow k=3,y=-1$

or,

$\Rightarrow 2k+y=35$

$\Rightarrow 2k-y=1$

$\Rightarrow k=9,y=17$

or

$\Rightarrow 2k+y=-35$

$\Rightarrow 2k-y=-1$

$\Rightarrow k=-9,y=-17$

or

$\Rightarrow 2k+y=-5$

$\Rightarrow 2k-y=-7$

$\Rightarrow k=-3,y=1$

So we have all the solutions.

Check for other cases also when the values gets reversed.

We have $n=(19+y)/2$

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Thanks Abhra Abir Kundu for giving me a precious solution. But I want to check using ordered pairs like $((2n-19),2k) = (\pm 6, \pm 1)$ my qusetion is how can i calculate other ordered pairs like this –  juantheron Mar 2 '13 at 8:54
You are welcome. @juantheron You can calculate the values of n using the formula I have just posted which will give u yours ordered pairs. –  Abhra Abir Kundu Mar 2 '13 at 8:57
The first equation is wrong ( I know, the OP wrote it): it must $\,-35\,$ instead... –  DonAntonio Mar 2 '13 at 14:01

Perhaps I'm missing something, but I think there's a basic mistake from the beginning:

$$n^2-19n+99=k^2\Longleftrightarrow4n^2-76n+396=4k^2\Longleftrightarrow (2n-19)^2\color{red}{+}35=(2k)^2\Longleftrightarrow$$

$$\Longleftrightarrow (2(n-k)-19)(2(n+k)-19)=-35$$

From here it follows at once that you need to express $\,-35\,$ as a product of odd integers with different sign , say for example:

$$2(n-k)-19=\pm 35\Longleftrightarrow n-k=\begin{cases}\;27\\{}\\\!-8\end{cases}\\{}\\2(n+k)-19=\mp 1 \Longleftrightarrow n+k=\begin{cases}\;\;9\\{}\\10{}\end{cases}$$

$$2n=\begin{cases}18\\\;\;2\end{cases}\;\;\Longrightarrow \;\;\text{two solutions,}\;\;n=9\,,\,1$$
Do the above again with $\,\pm 5\;\;,\;\;\mp 7\,$ for other possible solutions (hint: at least one solution more)