Find all integer solutions of m^3 - n^3 - 6mn = 8? What methods can I usually apply in solving this kind of Q?

Question

...in this kind of question?

My attempt :

First, either both m and n are odd, or both m and n are even.

m^3 - n^3 = (m-n)(m^2 + mn + n^2) = 8+6mn = 2(4+3mn)

(and i'm stuck. As you can see, I haven't got a clue.)

Answer 1

Nothing like trial and error and guesswork!!!
I presume only positive solutions for m and n are valid.
m^3 - n^3 - 6mn = 8

Rearrange to : m^3 = n^3 + 6mn + 8
Here it is seen that m > n, so let m = n + p,
noting that p > 0.

I usually try that first as it often gets rid of high powers.

Substituting this into the original equation gives :
(n + p)^3 - n^3 - 6n(n + p) = 8

Expanding gives :
n^3 + 3n^2p + 3np^2 + p^3 - n^3 - 6n^2 - 6np - 8 = 0

This gets rid of n^3, and we're left with a quadratic in n :
n^2(3p -6) + n(3p^2 - 6p) + p^3 - 8 = 0

Taking out factors and/or factorising gives :
3n^2(p - 2) + 3np(p - 2) + (p - 2)(p^2 + 4p + 2) = 0

Now take out the factor (p - 2) :
(p - 2)(3n^2 + 3np + p^2 + 4p + 2) = 0

Thus, there is a solution of p = 2.

Are there any more solutions?
No, because the other factor is always
positive, and so cannot be equal to zero.

Therefore the only solution is p = 2.
So m = n + 2.

This works for n >= 0.

Answer 2

If you have MATLAB of MAPLE. Plot the equation. This will give you all the points on this figure.
Also an C or Matlab code loop would be able to solve the problem. This is the easiest way I would do it.

You can download a C complier fairly simple. Go to Yahoo, and search for "Free C Complier". There are tons. Then it's just a matter if writing a small loop.

Answer 3

All solutions are given by m = t + 2 and n = t

Proof: Rearrange terms and factor to get

m(m^2-6n) = (n+2)(n^2 + 2n +4) then comparing factors

m<= (n+2) iff m^2-6n >= n^2 +2n +4 which is the same as

m<= n+2 iff m^2 >= (n+2)^2 or m >=(n+2) Therefore,

since the reverse condition holds as well we have m = n+2

QED

Answer 4

Equations of this type are called Diophantine Equations. I suggest you look for info on this in number theory books.

Answer 5

first of all for any variable to be found,number of equations must be equal to number of variables but in case of one equation and two variables,you can have infinite number of solutons and that can be solved by solving for instance (n)as a function of (m).that is all.

Answer 6

you cant, you only have 1 equation and two variables