Question

We are given four equations and we have to find n here:
\[4n = m^2 + k^2 + m + k\]
\[8056 = m^2 - k^2 + m - k\]
\[n - 2014 = \frac{k(k+1)}{2}\]
\[n + 2014 = \frac{m(m+1)}{2}\]

btw, I derived the first 2 equations from the bottom 2 equations.

Answer 1

@ganeshie8
@AravindG

Answer 2

@robtobey
@SolomonZelman

Answer 3

What were the original equations?

Answer 4

The bottom two.

Answer 5

its solvable :)

Answer 6

i got 2n=1
n=1/2

Answer 7

n has to be a natural number.

Answer 8

:/

Answer 9

wait :3

Answer 10

I got

2 x 2 x 2 x 19 x 53 = (m+k+1)(m-k)

Now (m+k+1) is always greater than (m-k)

We CAN use this to find 'n' but it'll take a heck lot of time!

Answer 11

sorrt idk lol
@ganeshie8 wanna try ?

Answer 12

@agent0smith @nincompoop

Answer 13

Idk, but to me it looks like you have to think of isolating one of the variables using "what 2014 is equal to", but other than that it is 2 equations with 3 variables.

Also knowing that it has to be positive integer... too much thinking for me, sorry -;(

Answer 14

its not like that lol , but i know nothing abt k,m how could i continue ?

Answer 15

whats the whol question
cuz i dnt read the horoscope lolz

Answer 16

Now (m+k+1) is always greater than (m-k) , only when m,k possitive

Answer 17

yes also : 2 x 2 x 2 x 19 x 53 = (m+k+1)(m-k)

you have only 16 cases to consider

Answer 18

so why do u think it gona take lot of time ?

and an observation - right hand side of both equations is a triangular number

Answer 19

But I'd have to solve for m and k, 16 times, and i may get n only few times.

m and k and n are all greater than zero.

Answer 20

ohh lol
so n-2014
n+2014
should apply

Answer 21

Also notice that if m-k is even, then m+k+1 has to be odd and vice versa

Answer 22

why?

Answer 23

oh got it ok.

Answer 24

so that reduces our work by more than a half :)

Answer 25

Now how many cases are going to be there?

Answer 26

the even prime powers : 2^3 must all lie in either m-k or m+k+1

Answer 27

m-k and m+k+1 cannot share 2's

Answer 28

Oh yeah!

So 8 * 19 * 53 are the options now right?

Answer 29

Looks good ! 8 cases to consider now

Answer 30

3 cases right? ;____;

8*19 * 53
19*53 * 8
8*53 * 19

Answer 31

@ganeshie8 ?

Answer 32

m-k = {1, 8,19, 53, 8*19, 19*53, 53*8, 8*19*53}

Answer 33

dont be so much relieved... :P there are 8 cases to work^^

Answer 34

Oh yeah, how could i forget the '1'.
Yeah, not relieved at all. ;__;

Answer 35

i dont see any easy ways out... u need to test all 8 cases i think... goodluck !

Answer 36

Thanks but, you helped me narrow it down to 8 cases. T___T

Answer 37

http://www.wolframalpha.com/input/?i=solve+in+integers+8056+%3D+%28m%2Bk%2B1%29%28m-k%29

Answer 38

use that to verify/reverse engineer ^

Answer 39

m and k both always have to positive right?
Because they form, the triangular numbers?

Answer 40

I think so, i never heard of negative triangular numbers... @BSwan do u ?

Answer 41

nope

Answer 42

http://www.wolframalpha.com/input/?i=solve+in+integers+8056+%3D+%28m%2Bk%2B1%29%28m-k%29%2C+m%3E0%2C+k%3E0