integer value of n

no. of integer values of $n$ for which $n^2+25n+19$ is a perfect square.

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

There are $6$ such integer values for which the value of the given polynomial is a perfect square.

Let $n^{2}+25n+19=m^{2}$ ,for some integer $m$ .

$=>n^{2}+25n+(19-m^{2})=0$ The discriminant on simplifying gives: $D=549+4m^{2}$ . Now, since $n$ is a perfect square,therefore $D$ has also to be a perfect square. Let, $549+4m^{2}=l^{2}$ ,for some integer $l$ . Therefore, $549=l^{2}-4m^{2}=(l-2m)(l+2m)$ . $549 = 183\times 3=549\times 1=9\times 61$ .Now,on putting $(l+2m)$ as $183,543$ and $61$ and $(l-2m)$ as $3,1$ and $9$ respectively and solving the respective equations gives us 3 different values of $m$ .On putting these values of $m$ in $\frac{-25 \pm \sqrt{549+4m^{2}}}{2}$ we get $6$ different values of $n$ for which the given polynomial is a perfect square.Also, when we put $l+2m = -9,-3,-1$ and $l-2m=-61,-183,-549$ respectively, we get the same values of $n$ as in the previous case.

Bhargav Das - 7 years, 8 months ago

The integer $n$ can be negative. Thus we also have $n=-30,-59,-150$ corresponding to $m=13,45,137$ , in addition to the answers you have given (put $\ell+2m = -9,-3,-1$ and $\ell-2m=-61,-183,-549$ respectively).

Mark Hennings - 7 years, 8 months ago

Corrected,thanks for correcting me.

Bhargav Das - 7 years, 8 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$