IMO2003 problem A2

[eepiccolo]

Here is the second problem from the 44th International Mathematical Olympiad. The other five are posted also. I don't know the answers, but I'll be working on them when I get the chance.

A2. Find all pairs (m, n) of positive integers such that m^2/(2mn^2 - n^3 + 1) is a positive integer

[wpolly]

Solution:

We Prove that the only solutions are (m,2m), here m is a positive integer.

Assume that [code:1]m^2/(2mn^2-n^3+1)=k[/code:1]
Thus we have
[code:1]m^2=k(2mn^2-n^3+1) (1)[/code:1]
Or, equivlently,
[code:1]m^2-k=kn^2(2m-n)[/code:1]
Mod m and we have
[code:1]-k=-kn^3 (mod m)[/code:1]
Thus[code:1]m|k(n^3-1)[/code:1]
So assume that [code:1]k(n^3-1)=lm (2)[/code:1]
Substitute (2) into (1) and then divide by m we have
[code:1]2kn^2=l+m (3)[/code:1]
Thus we see that if (m,n) is a solution then (l,n) is also a solution.
n*(3)-2*(2) we have that
[code:1](2l-n)(2m-n)=n^2-4k (4)[/code:1]

On the other hand, we have
[code:1]m^2>=n^2(2m-n)+1[/code:1]
n=2m is obvious a solution.
When 2m>n,we have 2m-n>=1.
So that[code:1](m/n)^2>=2m-n+1/n^2>=1+1/n^2>1[/code:1]
Thus m>n.
By the fact we mentioned above, we have l>n.
So 2m-n>n, 2l-n>n.
And finally we have that
[code:1]n^2-4k=(2l-n)(2m-n)>n^2[/code:1]
Which follows that k<0, CONTRADICTION.
So the only solutions are (m,2m).
Q.E.D.

[NickGlover]

(m, 2m) does cover some of the solutions, but there are others.

First, there is (m, 0).

Proof:

Substitute 0 for n into m^2/(2mn^2 - n^3 + 1) to get m^2/1 = m^2 which is obviously a positive integer for all integers, m.

Second, there is (2k, 1), where k is a positive integer.

Proof:

Substitute to get 4k^2/(4k - 1 + 1) = 4k^2/4k = k which is a positive integer by definition.

[NickGlover]

Oh, and a simpler proof for (m, 2m) is simply to substitute:

m^2/(2m*(2m)^2 - (2m)^3 + 1) = m^2/( (2m)^3 - (2m)^3 + 1) = m^2/1 = m^2

I don't immediately see how to prove there are no other solutions besides (m, 0), (2k, 1), and (m, 2m).

[eepiccolo]

Originally Posted by NickGlover

(m, 2m) does cover some of the solutions, but there are others.

First, there is (m, 0).

I don't think (m, 0) is valid because the integers are supposed to be positive, and 0 is neither positive nor negative.

[NickGlover]

Originally Posted by eepiccolo

I don't think (m, 0) is valid because the integers are supposed to be positive, and 0 is neither positive nor negative.

Oh, yeah, of course. I was...um...just testing you. :D

[Jwb52z]

How can 0 be a neutral number?

[rpresser]

Positive numbers are those that are greater than zero.
Negative numbers are those that are less than zero.
Zero is the additive identity element.

[hyh1048576]

Originally Posted by wpolly

We Prove that the only solutions are (m,2m), here m is a positive integer.

It seems to be right, but......

Originally Posted by wpolly aslo

Thus we see that if (m,n) is a solution then (l,n) is also a solution.

:? That's funny! The latter sentence shows that (8m^4-m,2m) where m is a positive integer is also solutions(that's because m+l=2kn^2=8m^4).

What's wrong?Here:

Originally Posted by wpolly

By the fact we mentioned above, we have l>n.

That's not right ;) .When m>n, l may be <n.

And I think the solutions are (m,2m), (8m^4-m,2m) and (2k,1) (m,k is positive integer)

[Edit:I have made a mistake-----It's (2k,1), not (1,2k) ops: ]
[Edit again: So the end of wpolly's proof should be modify like this:
n should be equal to one of 2m and 2l. If not so, n<2m and n<2l, so 0<(2m-n)(2l-n)=4ml-2n(m+l)+n^2=4k(n^3-1)-2n*2kn^2+n^2=n^2-4k
So we get n^2>4k>4m^2=(2m)^2>n^2.......CONTRADICTION :)

p.s. I don't think m>n and l>n is needed ;) ]

[NickGlover]

Originally Posted by hyh1048576

And I guess the solutions are (m,2m), (8m^4-m,2m) and (1,2k) (m is positive integer and k>1) :? :arrow:

It's not (1,2k), it's (2k,1), and k can be 1 or any other positive integer (the k>1 restriction is unnecessary).

Oh, and the simple proof of (8m^4-m,2m) is:

(8m^4-m)^2/(2(8m^4-m)(2m)^2 - (2m)^3 + 1)
= (64m^8-16m^5+m^2)/((8m^2)(8m^4-m) - 8m^3 + 1)
= (m^2)(64m^6 - 16m^3 + 1)/(64m^6 - 16m^3 + 1)
= m^2, which is obviously a positive integer since m is.