Finding Integers for a Fractional Equation

[harry654]

Good day!
I have problem:
Find all integers for which is fraction (n³+2010)/(n²+2010) equals to integer.

I can find 0 and 1 and I tried prove that any integers don't exist, but I didnt contrive it. Could someone help me with it?

 

[MathematicalPhysicist]

So you want (n^3+2010)=m(n^2+2010)
which is like saying: n^2 (n-m)= 2010(m-1)
which is the smae as: n^2= 2010(m-1)/(n-m)

One side is a square in order for the rhs to be a square too you have to get:
(m-1)/(n-m) = 2010^(2s+1)

or m-1 = t^2 and 2010/(n-m)=r^2

In both case I would rearrnage to get n as function of m and the other parameters, and check for the cases it can happen.
Also look that I divided by n-m if n=m then: m=1=n.

 

[harry654]

Thank you. So all integers for which is fraction equal to integer is 0 and 1?

 

[D H]

There is at least one more solution in the integers.

 

[Delta2]

Maybe this would help since we have to do with a cubic equation with a=1 b=-m c=0 d=-2010(m-1).

The discriminant in this case is $$8040m^3(m-1)-27(2010m-2010)^2$$. if descriminant is <0 then you only have to look for 1 real root n and check if it is integer whether m is integer.

 

[D H]

Please remember that we have a rule against doing other people's homework for them. The original poster saw post #2, so that damage has already been done. However, post #2 is not the complete answer. There is at least one other solution in the integers.

Please leave the rest as a problem for the original post. Do not solve it for him.

 

[harry654]

Solution: 0, 1, -2010

 

[Delta2]

Damn and i tried 2010 but didnt thought of -2010 lol.