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Calculate the integer value(s) of $n$ for which $^2 + 9 + 14$ is a perfect square.

Suppose $^2 + 9 + 14 = k^2$ where $k\in\mathbb{Z}$, reducing to the form $(2(n+k)+9) (2(n-k)+9) = 25$.

My question is how can I calculate for such integer values when the values I've gotten aren't integers

Thanks

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    Hint. Use the fact that the expression is (n+7)(n+2) – 2017-02-20

5 Answers 5

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We shall show that the given expression $n^2 + 9n + 14 =(n+2)(n+7)$ is a perfect square only for the case $\color{blue}{n=2}$ where $n\in\mathbb{N}$

Case 1

If both of these factors are squares, we only have $4$ and $9$ as squares which differ by $5$.

Case 2

If both of these factors are not perfect squares, then noting that $5 | (n+2,n+7)$, they are either coprime (in which case we go back to Case 1) or the only distributed factor among them is $5$. One have them has only one $5$, the other has $5^{2k+1}$, i.e. an odd number of $5$'s so as to make a complete square. So we can write one of them as $ 5^{2k+1}q^2$ and the other one as $5r^2$. Clearly, for the difference between them to be $5$, we will need $5^{2k}q^2 - r^2 = 1$, i.e. $(5^kq)^2 - r^2 = 1$ which is not possible since the difference between no 2 squares is 1 (unless one of them is zero)

Summary

We have shown that the expression $n^2 + 9n + 14 = (n+2)(n+7)$ is a perfect square only for the case $\color{blue}{n=2}$. Of course, when considering integers, you can also take $\color{blue}{n = -11}$ and get the same answer. The other trivial values (as pointed out by TonyK) are $\color{blue}{n = -2}$ and $\color{blue}{n = -7}$

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    You forgot $n=-2$ and $n=-7$. – 2017-02-20
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The given term is obviously not a perfect square for $n=1$ and for $n\geq 2$ we have:

$n^{2}+8n+16\leq n^{2}+9n+14

$\Rightarrow (n+4)^{2}\leq n^{2}+9n+14< (n+5)^{2}$

$\Rightarrow n^{2}+9n+14$ is a perfect square only for $n=2$ and is not a perfect square $\forall \, n\geq 3.$

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We have $$n^2+9n+14 = k^2$$ $$\Leftrightarrow (n+\frac {9}{2})^2 -\frac{25}{4}=k^2$$ $$\Leftrightarrow \frac {25}{4}=(n+\frac {9}{2})^2-k^2$$ $$\Leftrightarrow \frac {25}{4}=(k+n+\frac {9}{2})(n+\frac {9}{2}-k) $$

There are now only two choices left: $k+n+4.5 =0.5 \text { or } 12.5$ and $n+4.5-k = 12.5 \text { or } 0.5$.

Hope you can take it from here.

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$4(n^2 + 9n + 14) = (2n+9)^2-25$. Let $u=2n+9$.

So, if $v^2=n^2 + 9n + 14$ then $(2v)^2=4v^2=u^2-25$ and so $25=u^2-(2v)^2=(u-2v)(u+2v)$.

Now use that the only possible factorizations of $25$ are $1 \cdot 25$ and $5 \cdot 5$, except for order and sign. You must take both into account.

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Let $n^2+9n+14=(n+a)^2$ where $a$ is any integer

$\iff\dfrac{a^2-14}{9-2a}=n$ which is an integer

If integer $d>0$ divides both $a^2-14,9-2a;$

$d$ must divide $2(a^2-14)+a(9-2a)=9a-28$

$d$ must divide $9(9-2a)+2(9a-28)=25$

So integer $n,$ we need $9-2a$ must divide $25$

$\implies9-2a$ must lie $\in\{0,\pm1,\pm5,\pm25\}$

Check which values of $a$s keep $n$ an integer