2017 AIME II Problems/Problem 6

Revision as of 17:44, 25 November 2018 by Rootthreeovertwo (talk | contribs) (Solution 6)

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 (Using factors)

Let the expression be equal to $a$. This expression can be factored into $\sqrt{(n+44)^2-3n+81}$. 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 $71^2$, so $27$ is a solution. Proceed as Solution 5 to find the other solution(s).

See Also

2017 AIME II (ProblemsAnswer KeyResources)
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

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