2017 AIME II Problems/Problem 6

Revision as of 22:11, 28 January 2021 by Blehlivesonearth (talk | contribs) (Solution 4 (Bounding))

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 gives us the answer ; $168+27=\boxed{195}$.

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 (Squeezing/Sandwich method)

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).

Solution 7 (alternative factoring)

More intuitive, but a little bit slower considering the decimals.


Label the entire given expression as k^2.

Instinctively we can do a crude completion of the square, resulting in k^2 = (n+42.5)^2+210.75 Rearrange the equation to get a difference of squares.

k^2-(n+42.5)^2 = 210.75

(k+n+42.5)(k-n-42.5) = 210.75

Factor 21,075 to get 3^1,5^2, and 281^1

Now the two factors given are either divided by 10 each or one being divided by 100. Let's start with the former case.

If you try 281*3/10 and 5*5/10, you quickly realize that n becomes negative. Naturally, you will realize you want the number's difference to be larger. Try 281*5/10 and 3*5/10. This gives an answer of 27 for x. The next largest possibility also works, giving an n of 168. As you rise, some numbers don't work because it results in an n that is not an integer, as in the example of 281*5*5/10 and 3/10.

Now if you continue on with the next case, where one factor is divided by 100, very swiftly will you realize most don't work simply because the difference is too small, or it doesn't give an integer. It helps a lot when you realize that the decimal does not end in a 5, the answer will not be an integer. After a few short tests, we get $168+27=\boxed{195}$.


-jackshi2006

Video Solution

https://youtu.be/Z23Yz05eblY

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS