Revision as of 12:29, 25 March 2017

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.

@@ Line 11: / Line 11: @@
 Since <math>n</math> is positive, there are two cases, which are simple (luckily). Remembering that <math>x</math> is a positive integer, then <math>x^2-2017</math> and <math>85-2x</math> are either both positive or both negative. The smallest value for which <math>x^2>2017</math> 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 <math>x<45</math> (from the numerator) and <math>85-2x<0</math>, which means <math>x>42</math>. This only gives two solutions, <math>x=43, 44</math>. Plugging these into the expression for <math>n</math>, we find that they result in 27 and 168, which both satisfy the initial question. Therefore, the answer is <math>168+27=\boxed{195}</math>.
+==Solution 3 (Abuse the discriminant)==
+Let the integer given by the square root be represented by <math>x</math>. Then <math>0 = n^2 + 85n + 2017 - x^2</math>. For this to have rational solutions for <math>n</math> (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, <math>b^2 - 4ac = 7225 + 4x^2 - 8068 = y^2</math> for some integer <math>y</math>. Then <math>4x^2 - 843 = y^2</math>. Rearranging this equation yields that <math>843 = (2x+y)(2x-y)</math>. Noticing that there are 2 factor pairs of <math>843</math>, namely, <math>1*843</math> and <math>3*281</math>, there are 2 systems to solve for <math>x</math> and <math>y</math> that create rational <math>n</math>. These yield solutions <math>(x,y)</math> of <math>(211, 421)</math> and <math>(71, 139)</math>.
+The solution to the initial quadratic in <math>n</math> must then be <math>\frac{-85 \pm \sqrt{85^2 - 4(2017 - x^2)}}{2}</math>. Noticing that for each value of <math>x</math> that has rational solutions for <math>n</math>, the corresponding value of the square root of the discriminant is <math>y</math>, the formula can be rewritten as <math>n = \frac{-85 \pm y}{2}</math>. One solution is <math>\frac{421 - 85}{2} = 168</math> and the other solution is <math>\frac{139 - 85}{2} = 27</math>. Thus the answer is <math>168 + 27 = \boxed{195}</math> as both rational solutions are integers.
 =See Also=
 {{AIME box|year=2017|n=II|num-b=5|num-a=7}}
 {{MAA Notice}}

2017 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

Page

Toolbox

Search

Difference between revisions of "2017 AIME II Problems/Problem 6"

Revision as of 12:29, 25 March 2017

Contents

Problem

Solution 1

Solution 2

Solution 3 (Abuse the discriminant)

See Also