Once You Get Hold Of It, It's Easy-Part 3

Let $n^{2}+19n+92=x^{2}$ where $x$ is an integer. Then

$4x^{2}=4n^{2}+76n+368=(2n+19)^{2}+7$

Therefore,writing $A=2x$ and $B=2n+19$ we have $A^{2}-B^{2}=7$ , or $(A-B)(A+B))=7$ .Since $7$ is a prime number,it factorizes in just two ways:as $\pm1\times\pm7$ .Therefore either,

$2x-(2n+19)=1$ and $2x+(2n+19)=7$

leading to $4x=8$ and $4n+38=6$ or $x=2$ , $n=-8$ OR

$2x-(2n+19)=-1$ and $2x+(2n+19)=-7$

leading to $4x=-8$ and $4n+38=-6$ or $x=-2$ , $n=-11$ .

Thus,there are just two integers $n$ for which $n^{2}+19n+92$ is a square,namely $-8$ and $-11$ .Hence the answer $\boxed {-19}$

Completing the square: $(n+\frac{19}{2})^2 + \frac{7}{4} = x^2$ Multiplying by 4 and factorizing: $(2n+19)^2 = (2x)^2 - 7$ Now let ${m}$ = 2 ${n}$ and ${y}$ = 2 ${x}$ , such that: $(m+19)^2 = y^2 - 7 \Rightarrow\ (m+19)^2 = 9$ This means that: $m= -16 \ and \ m= -22 \Rightarrow\ n = -8 \ and \ n = -11 \therefore\sum_{}^{} n = -19$

Very easy and short.

$(n+9)(n+10) + 2$

is the product of two consecutive numbers, + 2.

So it is a perfect square if and only if the products of these two consecutive numbers equals $1 \cdot 2$ (as you can see it immediately via geometric interpretation).

This happens for $n = -8$ and $n = -11$ . The sum is then $\boxed{-19}$ .

In my solution all the variables apart from the original n are positive integers.

From the stated condition we can write

$n^2+19n+92=k^2$ which can be written as

$n^2+19n +(92-k^2)=0$ ........(1)

For each value of $k$ this formula gives two values of n which, by Vieta's formula, add to give $-19$ . All we need to do to complete the problem is find out how many values of k lead to integer solutions of (1), and multiply this number by $-19$ .

For integer solutions the discriminantof (1) must be a perfect square, so

$19^2-4(92-k^2)=p^2$

$\implies 4k^2-7=p^2$

$\implies (2k)^2-p^2=7$

$\implies (2k+p)(2k-p)=7$

This has exactly $\boxed{one}$ solution in positive integers, (it happens to be $k=3,p=1$ , but the precise values are not important for my solution!)

And so the sum of the roots is $1\times-19=\boxed{-19}$

Let $n^{2}+19n+92=s^{2}$

Now since $n \in I$ so Discriminant of the above $eq^{n}$ is whole square.

So $D=b^{2}-4ac = 19^{2}-4(92-s^{2})=k^{2}$

$7=4s^{2}-k^{2}=(2s+k).(2s-k)$ . Now we can have 2 equations

$2s-k=\pm1$

$2s+k=\pm7$

By this $s=\pm2 \Rightarrow \ s^{2}=4$ . Put this value in the given equation to get .

$n^{2}+19n+88=(n+11).(n+8)=0$

So $n=-8,-11$

By this the answer is $\huge-19$

$Let ~k~be~an~integer~ so~~~\displaystyle n^{2}+19n+92= k^2.$ $\displaystyle \text{So we have a quadratic equation }~n^2+19n+(92- k^2)=0.\\\text{For a perffect square, } 19^2-4*(92- k^2)=-7+k^2 \text{ must be a perfect square.} \\ \text{with integral k. k=4 satisfies this conditions. And the equation in unknown n }\\ \text{has by, Vieta's, the sum of the roots as -19}$