Find all pairs of integers (m,n) such that m+n=3(mn+10).
Problem
Source: KAMO 2022 Grades 7-8 P1
Tags: number theory, Diophantine equation
07.07.2022 10:52
m=n−33n−1∈Z⇒3n−93n−1=1−83n−1∈Z⇒3n−1∈{±1,±2,±4,±8}
07.04.2023 20:42
3m+3n-9mn=90 (1-3m)(3n-1)=89 By factors that is obvious
26.04.2023 22:03
m+n=3mn+30⇒3mn−m−n=−30⇒9mn−3m−3n=−90⇒9mn−3m−3n+1−1=−90 ⇒(3m−1)(3n−1)=−89, which leaves us with 4 cases: (i) 3m−1=−1,3n−1=89⇒m=0,n=30 (ii) 3m−1=89,3n−1=−1⇒m=30,n=0 (iii) 3m−1=1,3n−1=−89, no integer solutions (iv) 3m−1=−89,3n−1=1, no integer solutions Therefore, (m,n)=(0,30),(30,0) are the only integer solutions ◼ If only I did this when taking the test...
05.09.2023 03:11
Can't you use SFFT? 3mn−m−n=−30, so mn−m/3−n/3=−10, so (m-1/3)(n-1/3)=-89/9 then we multiply both sides by 9 i think or something like that to get (9m-3)(9n-3)=-89 wait.. isnt this wrong oops
17.06.2024 17:10
m+n=3(mn+10)⟺m−3mn=30−n m(1−3n)=30−n⟺m=30−n1−3n 3m=90−3n1−3n⟺3m=90−3n−1+3n1−3n=891−3n 3m∈Z⇒891−3n∈Z 1−3n∈{±1,±89} if we try this values, we see that: n∈{0,30} and put them in main equation (m,n)∈{(0,30),(30,0)} are solutions. We are done.