SOLUTION: Find couples m and n if mn-1|n^3-1. Numbers m and n are positive and elements of set N.

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: Find couples m and n if mn-1|n^3-1. Numbers m and n are positive and elements of set N.      Log On


   

Question 819841: Find couples m and n if mn-1|n^3-1. Numbers m and n are positive and elements of set N.
Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
(mn-1)|(n³-1)

Of course we must have n≧2 in all cases

If m=1, we have

(n-1)|(n³-1)

which of course is true, since n³-1 = (n-1)(n²+n+1)

If m=n², we have

(n³-1)|(n³-1)

which of course is true, since any natural number divides itself.

Now if n is a perfect square, say k², then we have

(mk²-1)|[(k²)³-1]

(mk²-1)|(k6-1)

Then we can have m=k, for then we'd have

(k³-1)|(k3-1)(k3+1)

So apparently there are 3 cases of solutions for n≧2

m=1, m=n², and the third case is when n is a perfect square, k², 
and m is its square root, k.

Edwin