Find all pairs (m, n) of positive integers such
that m²/(2mn² - n³ + 1) is a positive integer.
background of the problem ..some preliminary concepts and review of
related literature
m²
———————————————
2mn² - n³ + 1
Case 1. When the denominator = 1, then the above fraction is m², which
is a positive integer
The denominator = 1 is true when and only when
2mn² - n³ + 1 = 1, which is true when and only when
2mn² - n³ = 0, which is true when and only when
n²(2m - n) = 0
n is not 0 since it is a positive integer, so the last
equation is true when and only when
2m - n = 0, which is true when and only when
2m = n
So case 1 tells us that the given expression is a positive
integer when n = 2m. So the ordered pair (2m,m) will always
result in a positive integer for the given expression.
---------------------------------------
Case 2. When n = 1, the last two terms cancel
m²
——————————————— =
2mn² - n³ + 1
m²
——————————————————— =
2m(1)² - (1)³ + 1
m²
———— =
2m
m
———
2
And if m is chosen even, say m = 2k,
then the expression will be an integer,
So pairs (m,n) = (2k,1)
----------------------------------------
When n = 2
m²
——————————————— =
2mn² - n³ + 1
m²
——————————————————— =
2m(2)² - (2)³ + 1
m²
———————————— =
8m - 8 + 1
m²
———————— =
8m - 7
Divide numerator and denominator
by m
m
————————— =
8 - 7/m
This is an integer when m = 1 or 7, so
we have solutions (2,7) and (2,1). (2,1)
is a solution form case 1, so the only new
solution we have is (n,m) = (2,7)
My computer found another solution
(n,m) = (4, 126)
and I'm not sure how to get it. Sorry!
What math course is this? Number theory?
But apparently all the solutions are:
(2m,m), (2k,1), (2,7), and (4, 126)
where m, and k can be any positive integers.
Edwin
