1) gcd(m^3,n^2)=2^2*3^2
2) lcm(m^2,n^3)=2^4*3^4*5^6
from 1 we can deduce that both m,n are divisible by 2 and 3.
from 2 we know that one of them is divisible by 5, however it can't
be both because otherwise 1 would have a factor of 5.
let
m=2^a*3^b*5^i
n=2^c*3^d*5^j
where either i=3,j=0 or i=0,j=2 to reflect which one has the factor of 5
from 1 we get
min(3a,2c)=2
min(3b,2d)=2
since 3 does not divide 2 we know that the min value comes from 2c,2d thus
c=1,d=1
from 2 we get
max(2a,3)=4
max(2b,3)=4
so from this we can deduce a=b=2
thus we have 2 solutions
1) m=2^2*3^2*5^3 and n=2^1*3^1
2) m=2^2*3^2 and n=2^1*3^1*5^2

Posted by Daniel
on 20190503 20:12:45 