Aren't quadratics larger?

Number Theory Level 3

Find the sum of all (possibly negative) integers $n$ such that

$n^2+2 \mid 2014n+2.$

The answer is 2013.

3 solutions

Melanka Saroad
Aug 17, 2014

n^2+2 | (2014n+2)(2014n-2) then we get n^2+2 | 2*2014^2+4 but we know that 4 does not divide n^2+2 therefore n^2+2 | 2014^2+2. Now find the factors of 2014^2+2. Observe that the solutions of n obtained are such that n^2+2 | 2014n+2 or n^2+2 | 2014n-2 . So check validity of solutions. 0, 1, -2 and 2014 are the solutions.

Nice Solution and nice question @Finn Hulse :)

Happy Melodies - 6 years, 9 months ago

Aditya Raut
Aug 14, 2014

Once you start learning programming (and that too by self study), it does addict you, this is the way I used to solve the problem

img

Sum of all of them is $\boxed{2013}$

truly the unappreciated way, but that gave a practice of programming....

i want a mathematical solution not a CS program for this question

Rishabh Jain - 6 years, 10 months ago

See @Melanka Saroad 's solution. This problem has a standard approach.

Calvin Lin Staff - 6 years, 9 months ago

Can you justify that $n$ must be within that range?

Restricting the range of $n$ is a good way to approach this question.

Calvin Lin Staff - 6 years, 10 months ago

It's easy to see from the question that $n^2+2 \leq \left| 2014n+2 \right|$ , thus we get that for positive or $0$ value of $n$ , $n^2 \leq 2014n$ , and so for negative, It will be $n^2+2\leq -2014n-2$ , a quadratic $n^2+2014n+4\leq 0$

so the range is reduced. But as i had an idea of this range,to reduce those many calculations, I directly kept a big range, no matter, because the needed range is surely going to be in that :P

Aditya Raut - 6 years, 10 months ago

It is not true for n=3, if n=2015 how can $n^{2}$ +2 <=|2014n+2|?

shuvayan ghosh dastidar - 6 years, 9 months ago

@Shuvayan Ghosh Dastidar – Please! Bigger positives don't divide smaller numbers' absolute value ! WE ARE GIVEN $n^2+2 \mid 2014n +2$ and that does mean that $n^2+2 \leq 2014n +2$

Aditya Raut - 6 years, 9 months ago

Nice! I really love CS approaches, I'm an admirer of anybody who can program. :D

Finn Hulse - 6 years, 10 months ago

Is there a non-programming method?

Daniel Liu - 6 years, 10 months ago

Of course dude.

Finn Hulse - 6 years, 10 months ago

@Finn Hulse – I also added a note that $n$ may be negative because I usually don't see negative numbers with the $\mid$ sign.

Daniel Liu - 6 years, 10 months ago

I'm too lazy to add those integers. I too solved using Python. My code and output follows:

>>> li=[]
>>> for n in range(-1000000, 1000000):
    if ((2014*n + 2)%(n**2 + 2)==0):
        li.append(n)
        print sum(li)


-2
-2
-1
2013

So answer is 2013

Nafees Zakir - 6 years, 8 months ago

Carsten Meyer
Mar 31, 2021

Define $p(n):=n^2+2$ and $q(n):=2014n+2$ . Use Euclid's Algorithm to get possible common divisors: $\begin{aligned} 2\cdot 1007^2p(n) - (1007n-1)q(n) &= 4\cdot 1007^2+2=2\cdot 3\cdot 787 \cdot 859=:N &&&(*) \end{aligned}$ Let's apply " $\mod p(n)$ " on both sides of $(*)$ . We know $p(n)$ divides $q(n)$ , so the left part vanishes: $\begin{aligned} N &= 2\cdot 1007^2p(n)-(1007n-1)\equiv 0 \mod p(n) &\Rightarrow && p(n) \text{ divides }N \end{aligned}$ Notice $p(n)\geq 2$ must be a positive factor of $N$ - checking all $16$ of them we only find integer solutions for $\begin{array}{r|rrrr} p(n) & 2 & 3 & 6 & N\\\hline n & 0 & \pm 1 & \pm 2 & \pm 2014 \end{array}$ Find those $n$ that also satisfy " $q(n)\equiv \red{0}\mod p(n)$ ": $\begin{array}{r|rrrrrrrrrrr} n & 0 & 1 & -1 & 2 & -2 & 3 & -3 & 6 & -6 & 2014 & -2014\\\hline q(n)\mod p(n) & \red{0} & \red{0} & 1 & 4 & \red{0} & 5 & 10 & 2 & 2 & \red{0} & 4 \end{array}$ We have $n\in\{0;\:1;\:-2;\:2014\}$ , and the answer is $0+1-2+2014=\boxed{2013}$

Aren't quadratics larger?

The answer is 2013.

3 solutions

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