integer value of n

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

Note by Jagdish Singh
4 years, 8 months ago

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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 - 4 years, 8 months ago

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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 - 4 years, 8 months ago

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Corrected,thanks for correcting me.

Bhargav Das - 4 years, 8 months ago

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