Revision as of 02:08, 13 May 2018

Problem

Find the sum of all positive integers $n$ such that $\sqrt{n^2+85n+2017}$ is an integer.

Solution 1

Manipulating the given expression, $\sqrt{n^2+85n+2017}=\frac{1}{2}\sqrt{4n^2+340n+8068}=\frac{1}{2}\sqrt{(2n+85)^2+843}$ . The expression under the radical must be an square number for the entire expression to be an integer, so $(2n+85)^2+843=s^2$ . Rearranging, $s^2-(2n+85)^2=843$ . By difference of squares, $(s-(2n+85))(s+(2n+85))=1\times843=3\times281$ . It is easy to check that those are all the factor pairs of 843. Considering each factor pair separately, $2n+85$ is found to be $421$ and $139$ . The two values of $n$ that satisfy one of the equations are $168$ and $27$ . Summing these together, the answer is $168+27=\boxed{195}$ .

Solution 2

Clearly, the result when $n$ is plugged into the given expression is larger than $n$ itself. Let $x$ be the positive difference between that result and $n$ , so that $\sqrt{n^2+85n+2017}=n+x$ . Squaring both sides and canceling the $n^2$ terms gives $85n+2017=2xn+x^2$ . Combining like terms, $(85-2x)n=x^2-2017$ , so

$n=\frac{x^2-2017}{85-2x}.$

Since $n$ is positive, there are two cases, which are simple (luckily). Remembering that $x$ is a positive integer, then $x^2-2017$ and $85-2x$ are either both positive or both negative. The smallest value for which $x^2>2017$ is 45, which makes the denominator, and the entire expression, negative. Evaluating the other case where numerator and denominator are both negative, then we have that $x<45$ (from the numerator) and $85-2x<0$ , which means $x>42$ . This only gives two solutions, $x=43, 44$ . Plugging these into the expression for $n$ , we find that they result in 27 and 168, which both satisfy the initial question. Therefore, the answer is $168+27=\boxed{195}$ .

Solution 3 (Abuse the discriminant)

Let the integer given by the square root be represented by $x$ . Then $0 = n^2 + 85n + 2017 - x^2$ . For this to have rational solutions for $n$ (checking whether they are integers is done later), the discriminant of this quadratic must be a perfect square. (This can be easily shown using the quadratic formula.)

Thus, $b^2 - 4ac = 7225 + 4x^2 - 8068 = y^2$ for some integer $y$ . Then $4x^2 - 843 = y^2$ . Rearranging this equation yields that $843 = (2x+y)(2x-y)$ . Noticing that there are 2 factor pairs of $843$ , namely, $1*843$ and $3*281$ , there are 2 systems to solve for $x$ and $y$ that create rational $n$ . These yield solutions $(x,y)$ of $(211, 421)$ and $(71, 139)$ .

The solution to the initial quadratic in $n$ must then be $\frac{-85 \pm \sqrt{85^2 - 4(2017 - x^2)}}{2}$ . Noticing that for each value of $x$ that has rational solutions for $n$ , the corresponding value of the square root of the discriminant is $y$ , the formula can be rewritten as $n = \frac{-85 \pm y}{2}$ . One solution is $\frac{421 - 85}{2} = 168$ and the other solution is $\frac{139 - 85}{2} = 27$ . Thus the answer is $168 + 27 = \boxed{195}$ as both rational solutions are integers.

Solution 4

Notice that $(n+42)^2= n^2+84n+1764$ . Also note that $(n+45)^2= n^2+90n+2025$ . Thus, $(n+42)^2< n^2+85n+2017<(n+45)^2$ where $n^2+85n+2017$ is a perfect square. Hence, $n^2+85n+2017= (n+43)^2$ or $n^2+85n+2017= (n+44)^2.$ Solving the two equations yields the two solutions $n= 168, 27$ . Therefore, our answer is $\boxed{195}$ .

Solution 5 (Introducing a Variable)

Let the expression be equal to $a$ . This expression can be factored into $(n+44)^2-3n+81$ . Then square both sides, and the expression becomes $(n+44)^2-3n+81=a^2$ . We have a difference of two squares. Rearranging, we have $(n+44+a)(n+44-a)=3(n-27)$ . By inspection, the only possible values for $(n+44-a)$ are 0 and 1. When $(n+44-a)=0$ , we must have $n-27=0$ . Therefore, $27$ is a solution. When we have $(n+44-a)=1$ , so $n=a-43$ . Plugging this back to $(n+44+a)=3(n-27)$ (since $(n+44-a)=1$ ), we find that $a=211 \implies n=168$ . Thus, the answer is $27+168= \boxed{195}$ .

-RootThreeOverTwo

Solution 6

Ignore the square root for now. This expression can be factored into $(n+44)^2-3n+81$ . Just by inspection, when $n=27$ , the expression becomes $61^2$ , so $27$ is a solution. Proceed as Solution 5 to find the other solution(s).

@@ Line 24: / Line 24: @@
 Notice that <math>(n+42)^2= n^2+84n+1764</math>. Also note that <math>(n+45)^2= n^2+90n+2025</math>. Thus, <cmath>(n+42)^2< n^2+85n+2017<(n+45)^2</cmath> where <math>n^2+85n+2017</math> is a perfect square. Hence,<cmath>n^2+85n+2017= (n+43)^2</cmath> or <cmath>n^2+85n+2017= (n+44)^2.</cmath> Solving the two equations yields the two solutions <math>n= 168, 27</math>. Therefore, our answer is <math>\boxed{195}</math>.
-==Solution 5==
+==Solution 5 (Introducing a Variable)==
 Let the expression be equal to <math>a</math>. This expression can be factored into <math>(n+44)^2-3n+81</math>. Then square both sides, and the expression becomes <math>(n+44)^2-3n+81=a^2</math>. We have a difference of two squares. Rearranging, we have <math>(n+44+a)(n+44-a)=3(n-27)</math>. By inspection, the only possible values for <math>(n+44-a)</math> are 0 and 1. When <math>(n+44-a)=0</math>, we must have <math>n-27=0</math>. Therefore, <math>27</math> is a solution. When we have <math>(n+44-a)=1</math>, so <math>n=a-43</math>. Plugging this back to <math>(n+44+a)=3(n-27)</math> (since <math>(n+44-a)=1</math>), we find that <math>a=211 \implies n=168</math>. Thus, the answer is <math>27+168= \boxed{195}</math>.
 -RootThreeOverTwo
 ==Solution 6==

2017 AIME II (Problems • Answer Key • Resources)
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