IIT JEE , AIEEE Forum - Mathematics , Algebra

elastiboysai

Find all pairs (m, n) of positive integers such that m²/(2mn² - n³ + 1) is a positive integer.

hsbhatt

When m|n³-1 ?

anchitsaini

konichiwa2x

edited.

hsbhatt

So, m²/(2n²m-n³+1) = k

If n =1, then for m even, we have infinite solutions.

Case 1: m divides k. You can easily check that the only feasible case is n=1 which is already discussed above.

If m does not divide k,

m²-2n²km+k(n³-1) = 0

If this quadratic has integer solutions then they are factors of n³-1

Case2: Suppose m = n-1

Then k(n²+n+1)+(n-1) = 2n²k

or k = (n-1)/(n²-n-1). The only feasible soln here is n=2 so m = n-1 = 1.

Case 3: m = n³-1

Then k+(n³-1) = 2n²k

k = (n³-1)/2n²-1

Suppose n = 2p, then k = 8p³-1/(8p²-1) = p+(p-1)/(8p²-1) . So, p=1

i.e. n=2, m=7

Suppose n = 2p+1, then k = p+(4p²+5p)/(8p²+8p+1), which gives n=1, m=0 not feasible.

Case 4: m = n²+n+1

k(n-1)+n²+n+1 = 2n²k

or k = n²+n+1/(2n²-n+1). which yields no feasible cases.

Hence if n

1, the only solutions for (m,n) are (1,2) and (2,7)

PS: I am tempted to make a comment about the pointlessness of posting such problems in a JEE forum, but I wont.

anchitsaini

sir
but if m=r , n=2r
the denominator would become one and hence we would get a positive integer as a soln

konichiwa2x

The denominator is

, so

. If n = 1, then m must be even, in other words, we have the solution (m, n) = (2k, 1).

So assume n > 1. Put

. Then we have a quadratic equation for m, namely

. This has solutions

, where N is the positive square root of

. Since

, N is certainly real. But the sum and product of the roots are both positive, so both roots must be positive. The sum is an integer, so if one root is a positive integer, then so is the other.

The larger root

is greater than

, so the

. But note that if

, then since h > 0, we must have the denominator

smaller than the numerator and hence

. So for the smaller root we cannot have

. But

must be non-negative (since h is positive), so

for the smaller root. Hence

. Now

, so

. Thus n must be even. Put

and we get the solutions

and

.

Hence Solutions are :

ps. IMO-2003 Dont rate me. This is the official solution.

hsbhatt

I would certainly like to have a word with elasti about this

konichiwa2x

yeah, lets all boink him lol

http://www.kalva.demon.co.uk/imo/isoln/isoln032.html

vineetnegi

when n=0

m=any number

n=1

m= even numbers

anchitsaini

vineet first of all n cannot be 0 as it has to be a positive integer
secondly see the soln given above!!

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