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# integer value of n

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

Note by Jagdish Singh
3 years, 3 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. · 3 years, 3 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). · 3 years, 3 months ago