Martin N. wrote: Find all quadruples $(a, b, p, n)$ of positive integers, such that $p$ is a prime and \[a^3 + b^3 = p^n\mbox{.}\] (2nd Benelux Mathematical Olympiad 2010, Problem 4) it's seems not be a hard problem, well first assume that $a$ and $b$ are nonzero , let $a=\lambda p^m$ and let $b=qp^l$ where $\lambda$ and $q$ are not divisible by $p$ Thus the original equation is equivalent to $\lambda^3+q^3.p^{3l-3m}=p^{n-3m}$ but we must have $m=l$ because if not we'll have $p\mid \lambda$ and that' s not true , therefore we just need to work on the new equation $a^3+b^3=p^n$ where $a$ and $b$ are coprime with $p$ , now assume that $p>3$ then $(a+b)(a^2-ab+b^2)=p^n \Rightarrow a+b=p^r$ and $a^2-ab+b^2=p^k$ for some strictly positive integer $r$ and $k$ , and because $a^2-ab+b^2=(a+b)^2-3ab$ we get $p^k(p^{2r-k}-1)=3ab$ hence $p\mid 3ab$ and this is not true , so $p=3$ or $3^k(3^{2r-k}-1)=3ab \Rightarrow k=1$ therefore $3^{2r-1}-1=ab$ and $(a-b)^2+ab=3$ which implies that $a-b=\pm1$ and $ab=2$ and $r=1$ and so $(1,2,3,2)$ is a solution and also $(2,1,3,2)$ , if $p=2$ we can easily see that the only solution is $(1,1,2,1)$ because we get $k=0$ or $a^2-ab+b^2=1$ , since $2^k \not | 3ab$ and we finish the problem .

Zsigmondy's theorem works too (it is definitely overkill, but why not xD). If $a$ and $b$ aren't divisible by $p$, then there exists a prime divisor of $a^3+b^3$ which does not divide $a+b$ or $a^2+b^2$, with the exception $2^3+1^3=3^2$. Therefore, since $p$ divides $a+b$, $a=2$ and $b=1$ or vice versa. If $a$ and/or $b$ are divisible by $p$, we see that $v_p(a)=v_p(b)$ and consequently, we can reduce the problem to the case $p \not |ab$. @abdek: $a=0$ isn't possible, since they are positive

FelixD wrote: @abdek: $a=0$ isn't possible, since they are positive Yes !! I just considered that $0 \ge 0$ but I don't now if it's a convention or not because in our country we suppose that $\mathbb{N}=\{0,1,2,...\}$ while in other countries $0\notin \mathbb{N}$

In Austria, $0 \in \mathbb{N}$ too but positive is still not non-negative

FelixD wrote: In Austria, $0 \in \mathbb{N}$ too but positive is still not non-negative I have a trouble with my English, sorry ... I had a confusion beteween "positive" and "non-negative" ..Thank you

Lifting exponent works too (and is by far not that much of an overkill as Zsigmundy's). So for $p=2$ we can wlog take $a,b$ odd (or divide the equation by $2^{v_2(a)}$). But then $a^2-ab+b^2$ is odd so it must be $1$. for $p>2$ we know that $p \mid a+b$ so by LE we have \[v_p(a^3+b^3) = v_p(3) + v_p(a+b) \] and we can say that $a^2-ab+b^2 \in \{1,3\} $ which essentially kills the problem.

Let $gcd(a, b)=d$ then $d|p^n$ then $d$ is a power of $p$ Then let $a=xp^k$ and $b=yp^k$ where $x$ and $y$ are natural, $k$ is a non-negative integer and $gcd(x, y)=1$ Then $x^3+y^3=p^{n-3k}$ so let $m=n-3k$ then $m$ is a non-negative integer. This equation is symmetric so w.l.o.g. we can assume that $x\leq y$ $(x+y)(x^2-xy+y^2)=p^m$ $gcd(x+y, x)=gcd(x, y)=gcd(x+y, y)=1$ so $gcd(x+y, xy)=1$ So $gcd(x+y, x^2-xy+y^2)=gcd(x+y, 3xy)=gcd(x+y, 3)$ If $3\not|x+y$ then $x+y$ and $x^2-xy+y^2$ are coprime powers of $p$ Then $x+y>1$ so $x^2-xy+y^2=1$ Then $0\leq(x-y)^2=1-xy\leq 0$ then $x-y=0$ so $x=y=1$ since $x$ and $y$ are coprime. Then $p^m=2$ then $p=2$ and $m=1$ Otherwise $3|x+y$ then $3|x^2-xy+y^2$ and $p=3$ Then $\frac{x+y}{3}$ and $\frac{x^2-xy+y^2}{3}$ are coprime power of $3$ Then $\frac{x+y}{3}=1$ or $\frac{x^2-xy+y^2}{3}=1$ If the former then $x=1$, $y=2$ and $m=2$ If the latter then $(x-y)^2=3-xy$ then $xy=2$ (otherwise $x=y$) then $x=1$, $y=2$ and $m=2$ So $(x, y, p, m)=(1, 1, 2, 1); (1, 2, 3, 2); (2, 1, 3, 2)$ So $(a, b, p, n)=(2^k, 2^k, 2, 3k+1); (3^k, 2\cdot 3^k, 3, 3k+2); (2\cdot 3^k, 3^k, 3, 3k+2)$ where $k$ is a non-negative integer.

Posted before $p^n|(a+b)(a^2-ab+b^2)$.If $a=b,$we get infinitely many solutions with $p=2$.Now $p|a,b$ infinite descent. So $p|a^2+2ab+b^2,a^2-ab+b^2\implies p|3ab$.So $p=3$ Also see here: http://books.google.com/books?id=ED_4-bmi47gC&printsec=frontcover&dq=problems+from+around+the+world+2000-2001&source=bl&ots=oe23I8xOIc&sig=hciYMsCdLMmYRKE5pg4Ysq1j1wo&hl=en&ei=2tJ7TITAC8Ok4QaTgu2OBg&sa=X&oi=book_result&ct=result&resnum=3&ved=0CCMQ6AEwAg#v=onepage&q&f=false