WakeUp wrote: Find all integers $x$ and $y$ such that $x^3\pm y^3 =2001p$, where $p$ is prime. hint: $(x\pm{y})(x^2\mp{xy}+y^2)=3*23*29*p$

Is the question supposed to be understood like this - i.e. that $x^3+y^3=2001p$ and $x^3-y^3=2001p$ Or can the prime number be different in each case, i.e. that $x^3+y^3=2001p$ and $x^3-y^3=2001q$ where p,q are different primes ? In the first case there are no solutions, because adding the two simultaneous equations gives $x^3=2001p = 3.23.29p$ 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

I think we have 2 problems 1) $x^3+y^3=2001p$ and 2)$x^3-y^3=2001p$. If $(x,y)$ is solution first problem, then $(x,-y)$ is solution second problem. Therefore we can consider only one $x^3-y^3=(x-y)(x^3+xy+y^2)=d(d^2+3xy)=d(d^2+3yd+3y^2)=2001p$ Therefore $3|d\to 3|p\to p=3$ (p is same for both problem). Let $d=3m|9*23*29$. Then $y^2+3my+3m^2=\frac{23*29}{m}$ or $y=\frac{-3m\pm \sqrt{\frac{23*29}{m}-3m^2}}{2}.$ If $m=23,29$ $D<0$. If $m=1$ we have not integer solution.

WakeUp wrote: Find all integers $x$ and $y$ such that $x^3\pm y^3 =2001p$, where $p$ is prime. Let prove that $2001$ divides $x,y$.We have $x^3 + y^3 = 2001p,x^3 - y^3 = 2001q \implies x^3 + y^3 = 0$ $mod$ $2001,x^3 - y^3 =0$ $mod$ $2001$ $\implies x^3 = 0$ $mod$ $2001 \implies x = 0$ $mod$ $2001 \implies y = 0$ $mod$ $2001 \implies x = 2001k,y = 2001l \implies 2001 ^ 2 * (k ^ 3 - l ^ 3) = q$,contradiction.There aren't any integer x,y that satisfies the egality.

You cant assume x^3==0 mod 2001 from x^3==y^3 mod 2001

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