Find all integers x and y such that x3±y3=2001p, where p is prime.
Problem
Source:
Tags: number theory, number theory proposed
30.10.2010 22:39
WakeUp wrote: Find all integers x and y such that x3±y3=2001p, where p is prime. hint: (x±y)(x2∓xy+y2)=3∗23∗29∗p
10.11.2010 10:10
Is the question supposed to be understood like this - i.e. that x3+y3=2001p and x3−y3=2001p Or can the prime number be different in each case, i.e. that x3+y3=2001p and x3−y3=2001q where p,q are different primes ? In the first case there are no solutions, because adding the two simultaneous equations gives x3=2001p=3.23.29p which obviously cant be a cube if p prime. In the second case, I think going mod 3 and using fermat's little theorem shows both x and y must be zero mod 3, and hence no solutions - but someone might like to check this. merlin
10.11.2010 12:27
I think we have 2 problems 1) x3+y3=2001p and 2)x3−y3=2001p. If (x,y) is solution first problem, then (x,−y) is solution second problem. Therefore we can consider only one x3−y3=(x−y)(x3+xy+y2)=d(d2+3xy)=d(d2+3yd+3y2)=2001p Therefore 3|d→3|p→p=3 (p is same for both problem). Let d=3m|9∗23∗29. Then y2+3my+3m2=23∗29m or y=−3m±√23∗29m−3m22. If m=23,29 D<0. If m=1 we have not integer solution.
13.05.2015 15:09
WakeUp wrote: Find all integers x and y such that x3±y3=2001p, where p is prime. Let prove that 2001 divides x,y.We have x3+y3=2001p,x3−y3=2001q⟹x3+y3=0 mod 2001,x3−y3=0 mod 2001 ⟹x3=0 mod 2001⟹x=0 mod 2001⟹y=0 mod 2001⟹x=2001k,y=2001l⟹20012∗(k3−l3)=q,contradiction.There aren't any integer x,y that satisfies the egality.
23.08.2020 12:02
You cant assume x^3==0 mod 2001 from x^3==y^3 mod 2001
04.04.2023 15:20
This is an easy question.Just notice that p=3 because if we factor the difference or sum of cubes, its clear that if one side is divisible by 3 then both sides should be divisible by 3. And then just factorise it