Ahiles wrote: We have $ p^2|a^3 + b^3 = (a + b)(a^2 + b^2 - ab)$ Because $ p$ is prime, there are 2 cases: 1) $ p|a + b$, so $ \boxed{p^2|a + b}$. Why from $ p|a+b$ then $ p^2|(a+b)$ We can general problem as follow : Problem $ p$ is a prime satisfy : $ p|a+b$ and $ p^k|a^n+b^n$ where $ n\equiv 1 (\mod 2)$ Prove that $ p^{k-v_p(n)}|a+b$ or $ p^n|a^n+b^n$ It is an consequence of this lemma : Lemma If $ \gcd(a,b)=1$ and $ p|a-b$ then $ v_p(a^n-b^n)=v_p(a-b)+v_p(n)$ It is an well know lemma .

TTsphn, we have $ p^2|(a + b)(a^2 - ab + b^2)$. Because $ p$ divides $ a + b$, it should divide or $ a + b$ one more time (so $ p^2|a + b$) or it divides $ a^2 - ab + b^2$

Ahiles wrote: TTsphn, we have $ p^2|(a + b)(a^2 - ab + b^2)$. Because $ p$ divides $ a + b$, it should divide or $ a + b$ one more time (so $ p^2|a + b$) or it divides $ a^2 - ab + b^2$ If that you should type : $ p|\frac{a+b}{p}$

Ahiles wrote: Let $ p$ be a prime number, $ p\not = 3$, and integers $ a,b$ such that $ p|a + b$ and $ p^2|a^3 + b^3$. Prove that $ p^2\mid a + b$ or $ p^3\mid a^3 + b^3$. (Lemma) If $ p|\frac{a^3+b^3}{a+b}$, then $ p^2|\frac{a^3+b^3}{a+b}$. Proof: Suppose the "if" statement. Clearly $ p|p^2|(a+b)^2$, so $ p|3ab$, so at least one of $ a,b$ is a multiple of $ p$. But then the other must be, also, because of $ p|a+b$. Then we have the stronger statement $ p^2|3ab$. Reverting back to the stronger $ p^2|(a+b)^2$, we deduce that $ p^2|\frac{a^3+b^3}{a+b}$. Using TTsphn's notation, we have two cases: 1. $ v_p(a+b)=1$ We compute: \[ v_p(a+b)+v_p\left(\frac{a^3+b^3}{a+b}\right) = v_p(a^3+b^3)\geq2\iff v_p\left(\frac{a^3+b^3}{a+b}\right)\geq1\] But all the lemma says is that $ v_p\left(\frac{a^3+b^3}{a+b}\right)\not=1$. Thus it is $ \geq2$; thus $ \boxed{v_p(a^3+b^3)\geq3}$. 2. $ v_p(a+b)>1$ Clearly this is equivalent to the other "or" case, so the problem is solved. TTsphn, where is a proof of that lemma? And i'm assuming it holds for negative $ b$? Because then the solution to the general case is really just arithmetic

Yes, of course it holds also for negative $ b$. Try to prove it yourself, it's only induction and binomial theorem.

What's the source of the problem?

Ahiles wrote: Let $ p$ be a prime number, $ p\not = 3$, and integers $ a,b$ such that $ p|a + b$ and $ p^2|a^3 + b^3$. Prove that $ p^2\mid a + b$ or $ p^3\mid a^3 + b^3$. $p|a+b,p^2|(a+b)^3-3ab(a+b)\implies p^2|3ab(a+b)\implies p|3ab\implies p|a,b\implies p^3|a^3+b^3$ Else we have $p\not|a^2-ab+b^2$ and then from Lifting Exponent Lemma,since $p|a+b,p^2||a^3+b^3$

It has been almost three years since my last post here, and I've learned some math since then. I can see that I once viewed this as an isolated, number theory problem. In fact, this is a very trivial problem in: http://en.wikipedia.org/wiki/P-adic_analysis

I don't see where you want to throw serious analysis in Well, one can do it using logarithms (I ignore $p=2$): If $p|b$, done. Otherwise, $p$-adically, you can reduce this to having some $\zeta = - \frac ab \in U^{(1)}$ such that $\zeta^3 \in U^{(2)}$. Thus if $\xi = \log_p(\zeta)$, you get $\xi \in \mathfrak p$ and $3 \xi \in \mathfrak p^2$, which means that $\xi \in \mathfrak p^2$ or $\zeta \in U^{(2)}$. Alternatively, you could directly involve the multiplicative structure of the $p$-adics, but that's just another use of $\log_p$.

$p^2|a^3+b^3$ if $p^2|(a+b)$ then done else $p|a^2-ab+b^2$ but, $p|(a+b)$ so, $p|(a+b)^2$ so, $p|3ab$ so, $p|ab$ since $p\neq 3$ let wlog $p|a$ then as $p|a+b$ so, $p|b$ so, $p^3|a^3+b^3$

kunny wrote: What's the source of the problem? This is Romanian Junior Balkan Team Selection Test 2008.

Thanks a lot.

$p$ divides $(a+b).$ so , $V_p{(a+b)}\ge1$ [by $V_p{(x)}$ , we mean the highest power of $p$ that divides $x$]. now , if $V_p{(a+b)}\ge2$ , then $p^2$ divides $(a+b)$ otherwise , suppose $V_p{(a+b)}=1$. so,$a+b=pk$ , where $p$ doesn't divide $k$ so , $a^3+b^3=(a+b)(a^2-ab+b^2)=pk(a^2-ab+b^2)$ so , $p$ divides $(a^2-ab+b^2)=(a+b)^2-3ab$. so , $p$ divides $3ab$. as $p$ is not equal to $3$ , we have, $p$ divides $ab$. so , $p$ divides $a$ or,$b$ or,both. now , as $p$ divides $(a+b)$ also , so $p$ divides both $a$ & $b$. so , $p^3$ divides $a^3+b^3$. hence done!

Why not this completely elementary approach? Write $d=\gcd(a,b)$, $a=dx$, $b=dy$, $\gcd(x,y)=1$. We are given $p\mid a+b = d(x+y)$ and $p^2 \mid a^3+b^3 = d^3(x^3+y^3)$ for some prime $p\neq 3$. Now, if $p\mid d$, it immediately follows $p^3 \mid d^3 \mid d^3(x^3+y^3) = a^3+b^3$ (it could be that $p^2 \nmid d(x+y)=a+b$, if $p^2\nmid d$ and $p\nmid x+y$). If, on the other hand, $p\nmid d$, then $p\mid x+y$. Assuming $p^2 \nmid x+y$, notice that, since $p^2 \mid x^3+y^3 = (x+y)^3 - 3xy(x+y)$ and $p\neq 3$, it follows $p\mid xy$; but this is inconsistent with $p\mid x+y$. Therefore it must be $p^2\mid x+y \mid a+b$ (and it could be that $p^3 \nmid d(x^3+y^3) = a^3+b^3$, if $p^3\nmid x+y$). Of course, a trivial case is for $0 = a+b = a^3+b^3$, when $p$ at any power divides both.

I was trying to write a one-line solution using LTE for this one, but I just realized this can't be done (because we finally show that $p | a$ and $p| b$)...

Why doesn't LTE work? Case 1: $a \not\equiv 0 \bmod{p}$ We then have by LTE that $v_p(a^3+b^3)= v_p(a+b)$. This automatically implies that $p^2|a+b$ by the conditions. Case 2: $a \equiv 0 \bmod{p}$ It is then easy to see that $p^3|a^3+b^3$ from the conditions. This is basically $2$ one liners.

if a=5 b=5 p=5 =>p*p|a+b this is wrong

samaridin wrote: if a=5 b=5 p=5 =>p*p|a+b this is wrong

p|(a+b) we have (a+b)/p€N and used p2|(a+b)(a2-ab+b2) we deduce that p| (a+b)/p(a2-ab+b2), (p2=p*p,a2=a*a ,b2=b*b,(a+b)2=(a+b)*(a+b) ) Because p is a prime number ,we have: a)p| (a+b)/p →p2|(a+b) b)p|(a2-ab+b2)→p|[(a+b)2-3*a*b] and p|(a+b)2 →p|3*a*b→p|a or p|b because p is a prime number p≠3 p|a+b→p|a and p|b→ p|a3 and p|b3→p|a3+b3 (a3=a*a*a and b3=b*b*b) In conclusion p|a+b or p|a3+b3