Possible Perfect Squares

Suppose $n^2+19n+130=k^2$ for some positive integer $k$ . Completing the square on the left hand side, $\left(n+\frac{19}{2}\right)^2+\frac{159}{4}=k^2$ . Multiplying by four to clear out the fractions, $(2n+19)^2+159=(2k)^2$

Now, this implies $(2k)^2-(2n+19)^2=159$ , or $(2k-2n-19)(2k+2n+19)=159=3\cdot53$ . Since $2k-2n-19<2k+2n+19$ , we have two cases:

In the first case, $2k-2n-19=1$ and $2k+2n+19=159$ . Adding these together, $4k=160$ , so $k=40$ and $n=30$ . We can see that this does indeed produce a perfect square.

In the second case, $2k-2n-19=3$ and $2k+2n+19=53$ . Similarly as above, we find that $k=14$ and $n=3$ , which again produces a perfect square.

Now we have the values of $n$ as $3$ and $30$ , so the sum of the values is $33$ .

Since for all positive $n$ , $(n+9)^2=n^2+18n+81<n^2+19n+130<n^2+24n+144=(n+12)^2$ , we have that $n^2+19n+130$ must be either $(n+10)^2$ or $(n+11)^2$ . In the first case, $n^2+19n+130=n^2+20n+100$ , and this yields $n=30$ . In the second case, $n^2+19n+130=n^2+22n+121$ , and this yield $n=3$ . Hence, the answer is $30+3=33$ .

let n^2+19n+130 = a^2

==>4n^2 + 76n + 520 = 4a^2

==>(2n + 19) ^ 2 + 159 = (2a)^2

==>(2a)^2 - (2n + 19)^2 = 159

==>(2a + 2n + 19)(2a - 2n - 19) = 159.

let x = 2a + 2n + 19, y = 2a - 2n - 19,

x - y = 4n + 38,

Cases:

1.) x = 159, y = 1 ==> n = 30

2.) x = 53, y = 3 ==> n = 3

3.) x = 3, y = 53 ==> n = -22

4.) x = 1, y = 159 ==> n = -49

5.) x = -159, y = -1 ==> n = -49

6.) x = -53, y = -3 ==> n = -22

7.) x = -3, y = -53 ==> n = 3

8.) x = -1, y = -159 ==> n = 30

Since we're only looking for positive integers n, n = 30, 3

sum of possible positive integer values of n = 30 + 3 = 33

We need to find all positive integral values of n for which given equation is a perfect square.

1) Let n^2 + 19n + 130 = p^2

2) 4n^2 + 2 x 2n x 19 + 4 x 130 = 4 x p^2 (multiply equation by 4)

3) 4n^2 + 2 x 2n x 19 + 361 - 361 + 520 = 4 x p^2 (add and subtract 361 in lhs)

4) (2n + 19)^2 + 159 = (2p)^2

5) (2p)^2 - (2n + 19)^2 = 159

6) (2p - 2n - 19) x (2p + 2n + 19) = 159

Now divisors of 159 are 1,3,53,159 159 can be expressed as either 159 = 1 x 159 or 159 = 3 x 53

7)

 a)  (2p - 2n - 19) x (2p + 2n + 19) = 1 x 159 
        So (2p - 2n - 19) = 1 and (2p + 2n + 19) =159
        On adding 4p = 160 or p =40 
        Giving n = 3 or -22 (we take positive value n = 3)

b)  (2p - 2n - 19) x (2p + 2n + 19) = 3 x 53
        So (2p - 2n - 19) = 3 and (2p + 2n + 19) =53
        On adding 4p = 56 or p =14 
        Giving n = 30 or -49 (we take positive value n = 30)

Hence sum of all possible positive integer values of n is 3 + 30 = 33

Our final answer is 33

Let n^2+19n+130=a^2. By completing the square, (n+19/2)^2+159/4=a^2. (2n+19)^2+159=(2a)^2, (2a)^2-(2n+19)^2=159. Factorising, (2a+2n+19)(2a-2n-19)=159. -----[I] Since both 2a+2n+19 and 159 are positive, 2a-2n-19 is also positive. Since both 2a+2n+19 and 2a-2n-19 are positive integers, write 159 as product of two positive integers. Since the prime factorization of 159 is 3x53, 159 can be written as the product of two positive integers in exactly two ways, that is 159x1 or 53x3. Since 2a+2n+19>2a-2n-19, by equating the factors of both sides of [I], 2a+2n+19=159 and 2a-2n-19=1 or 2a+2n+19=53 and 2a-2n-19=3. By solving simultaneous equations, a=40 when n=30, a=14 when n=3. By substituting the values of n into the given equation, both n=30 and n=3 satisfy all the condition of the problem. Therefore, the sum of all possible positive integer values of n is 30+3=33.

As $n^2+19n+130>n^2$ , we infer that $n^2+19n+130=(n+a)^2$ , where $a$ is positive. This means that $19n+139=a^2+2na$ , or $139-a^2=n(2a-19)$ and so either $a<10$ and $139<a^2$ , or $a^2<139$ and $a>9$ . It follows easily that the possible values of $a$ are only 10 and 11, giving $n=30$ and $n=3$ respectively. The sum of the possible values of $n$ is thus 33.

Consider $(n+12)^2$ and $(n+9)^2 = n^2 + 18n + 81$ . We will show that these are an upper and lower bound, respectively. $(n+11)^2 = n^2 + 24n + 144 > n^2+ 19n + 130 \Leftrightarrow 3n > -14$ , which is clearly true since $n$ is positive. $(n+9)^2 = n^2 + 18n + 81 < n^2 + 19n + 130 \Leftrightarrow n < -49$ , which is also true by the same reasoning. Therefore, the only possibilities are $n^2 + 19n + 130 = (n+10)^2$ or $(n+11)^2$ . Solving for both possibilities, we get two solutions: $n = 3$ and $n = 30$ . Summing the two, we obtain a final answer of $33$ .

$m \in \mathbb{Z} \Rightarrow n^2+19n+130 = m^2$

$n^2+19n+130-m^2=0$

$n = \frac{-19 + {(19^2-4(130-m^2))}^{\frac{1} {2}} }{2}$

$n = \frac{-19 + {(4m^2-159)}^{\frac{1} {2}} }{2}$

The term inside the root is a perfect square (since n is integer):

$4m^2-159 = k^2, k \in \mathbb{Z}$

$(2m-k)(2m+k) = 159$ $(*)\Rightarrow$ $(2m-k = 1\quad and \quad 2m+k = 159)$ $(2m-k = 3 \quad and \quad 2m+k = 53)$

$\Leftrightarrow$ $(m = 40 \quad and \quad k = 79)$ $(m = 14 \quad and \quad k = 25)$ $\Leftrightarrow$ $(n = 30)$ $(n = 3)$

Then the sum of all possible positive integer values of n is:

$3 + 30 = 33$

$(*)$ The negative solutions were neglected since n is a function of $m^2$ .

Suppose we have this equation

$n^2 + 19 n + 130 - k^2 = 0$ and $n > 0$ and $k \geq 0$

We want to find all of its integer solutions. If we now solve this equation for n, we will get two solutions, one with a positive sign in front of the square root and one with a negative sign. We will only observe the solution with the positive sign because the other one will surely be negative. The solution is

$\frac{1}{2}(\sqrt{4 k^2 - 159} - 19)$

Since $\sqrt{4 k^2 - 159}$ has to be an integer, $4 k^2 - 159$ has to be a perfect square. Observe the following: the gap between two consecutive perfect squares increases as we count them. Since $4 k^2$ is a perfect square, $4 k^2 - 159$ surely cannot be a perfect square starting from some value of k. If we take $k=41$ , the gap between $4 k^2$ and the previous perfect square, $(2k-1)^2$ will be equal to 163, and thus no value of k greater than 40 can be a solution. If we manually check all the possible values of k we will only get two pairs of valid solutions for $(n,k)$ : $(3,14)$ and $(30,40)$ . Thus, the sum of possible values for n is 33.

Completing the square we get that $n^2 + 19n + 130 = (\frac {2n+19}{2})^2 + \frac {159}{4}$

We know this must equal a square, which we will call $j^2$ .

So we equate: $j^2=(\frac {2n+19}{2})^2 + \frac {159}{4}$

Multiplying both sides by 4 and rearranging we have that $-159=( 2n+19)^2 - 4j^2$

Factoring by difference of squares we see that $-159= (2n+2j+19)(2n-2j+19)$

The two terms on the RHS can be any factor pair of $-159$ , however because their sum is $4n+38$ , we know that the two factors must sum to a number that is positive and 2 mod 4. The only factor pairs that satisfy this condition are $(159, -1)$ and $(53, -3)$ . In both cases we still get a solution for $j$ , so they do indeed work.

They produce values of $3$ and $30$ for $n$ so our answer is $3+30=\boxed{33}$

n^2 + 19n + 130 must be of the form n^2 + 2nx + x^2

=> x^2 + 2nx = 19n + 130 => n = (x^2 - 130) / (19 - 2x)

n is +ve if,

1) Both numerator and denominator are +ve.... or

2) Both numerator and denominator are -ve.....

1) If both are +ve ,

-----> x^2 > 130 , x > 11.4 ; -----> 19 > 2x , x < 9.5 => No solution.

2) If both are -ve ,

----->x^2 < 130 => x < 11.4 ; ----->19 < 2x , x > 9.5.

=> x = either 11 or 10 => n = 3 or 30

therefore , sum = 30 + 3 = 33

Let n^2+19n+130 = x^2, where x,n are positive integers. 4n^2+76n+520 = 4x^2 (2n+19)^2+159 = 4x^2 (2x)^2 - (2n+19)^2 = 159 = 3*53 (2x-2n-19)(2x+2n+19) = 159 Possible combinations are (1)(159) and (3)(53). Solving for x and n we get (x,n) = (40,30) and (14,3)

Let the perfect square equal $k^2$ . We start by completing the square

$n^2 + 19n + 130 = k^2$

$n^2 + 19n + (\frac{19}{2})^2 - \frac{19}{2})^2 + 130 = k^2$

$n^2 + \frac{19}{2})^2 + \frac{159}{4} = k^2$

Multiplying both sides of our equation by 4 will make things easier.

$(2n^2 + 19)^2 + 159 = 4k^2$

$159 = 4k^2 - (2n^2 + 19)^2$

$159 = (2k - 2n - 19)(2k + 2n + 19)$

$159$ has prime factorization $3 \cdot 53$ . Due to the fact that $n$ and $k$ are integers, $2k - 2n - 19$ and $2k + 2n + 19$ by closure. Also, note that $2k + 2n + 19 > 2k - 2n - 19$ because $n$ is a positive integer $\rightarrow 4n + 38 > 0 \forall n \in \mathbb{N}$ . So, we can compare our two terms with factorization of 159 into two integers $a \cdot b$ with $a<b$ ,

$a = 2k - 2n - 19$

$b = 2k + 2n + 19$

Case 1: $a = -159, b = -1$

Algebra gives the solution $k = -40, n = 30$ .

Case 2: $a = -53, b = -3$

Algebra gives the solution $k = -14, n = 3$ .

Case 3: $a = 1, b = 159$

Algebra gives the solution $k = 40, n = 30$ .

Case 4: $a = 3, b = 53$

Algebra gives the solution $k = 14, n = 3$ .

The answer is then $3 + 30 = 33$

Note that $n^2 + 19n + 130$ being a perfect square means that $n^2 + 19n + 130 = k^2$ where $k$ is an integer. Now, complete the square on the Left-hand-side:

$\left(n+\dfrac{19}{2}\right)^2+\dfrac{159}{4}=k^2$

Naturally, we multiply both sides of this equation by $4$ to get:

$\left(n+19\right)^2+159=4k^2$

Now, multiply both sides by $4$ , subtract the $n^2$ from both sides, and factored the RHS. We get:

$\left(2\cdot k+2\cdot n+19\right)\cdot\left(2\cdot k-2\cdot n-19\right)=159$

Prime-factorizing $159$ , we get:

$\left(2\cdot k+2\cdot n+19\right)\cdot\left(2\cdot k-2\cdot n-19\right)=53 \cdot 3$

If $2k + 2n + 19 = 53$ and $2k - 2n - 19 = 3$ , we add to get $4k = 16 \Rightarrow k = 14$ . We see that this means that $n = 3$ (just substitute into the original equation). This is a solution.

If $2k - 2n - 19 = 1$ and $2k + 2n + 19 = 159$ , we add to get $4k = 160$ . We see that this means that $n = 30$ . Again, just substitute to see this.

Since $n$ is positive and both $k=p$ and $k=-p$ are solutions (which means we only need to care about positive $k$ s), we remark that $2k - 2n - 19 < 2k + 2n + 19$ . So we know that we haven't missed anything.

So the only two results for $n$ were $3$ and $30$ , making the answer $3 + 30 = \boxed {33}$ .

Since $n$ is positive, $(n+9)^2 = n^2 + 18 n + 81 < n^2 + 19 n + 130 < n^2 +24 n + 144 = (n+12)^2$ . Since $n^2 + 19n + 130$ is a perfect square, we must have $n^2 + 19n + 130 = (n+10)^2 = n^2 + 20n + 100 \Rightarrow n = 30$ or $n^2 + 19n + 130 = (n+11)^2 = n^2 + 22n + 121 \Rightarrow n = 3$ . Hence, the sum of all possible values of $n$ is $3 + 30 = 33$ .

n^2+19n+130 can be written as (n+10)^2+30-n. Clearly n=30 gives a perfect square. By substituting n=3 we see that the expression becomes 〖13〗^2+30-3=196 which is a perfect square‼! Above n=30 the expression deviates away from a perfect square. The deviation is so small that it does not reach to the nearest left perfect square. ∴Sum of the numbers n so that it is a perfect square is 3+30=33