Consider a square-free even integer n and a prime p, such that 1) (n,p)=1; 2) p≤2√n; 3) There exists an integer k such that p|n+k2. Prove that there exists pairwise distinct positive integers a,b,c such that n=ab+bc+ca. Proposed by Hongbing Yu
Problem
Source: 2012cmo,problem5
Tags: modular arithmetic, inequalities, number theory proposed, number theory
12.01.2012 17:54
Sorry, but what is x?
13.01.2012 05:59
I think x should be replaced by n. I think this solution works (but I may have made a mistake). Solution Let k \equiv m \pmod{p} where 0 \le m < p. Note that m>0 since p|k would imply that p|n which is a contradiction. Since \gcd(n,p)=1 and n is even, it follows that p is odd. Hence p-m and m are of different parity. Let c be the odd positive integer in the set \{m, p-m \}. Since 0<m<p and p| n+k^2, it follows that c>0 and that p| n+c^2. Now let pq = n+c^2 where q\in \mathbb{N}. Now note that by AM-GM, q=\frac{n+c^2}{p} \ge \frac{2c\sqrt{n}}{p} \ge c. However, since n is square-free, it cannot follow that n=c^2 and therefore the inequality is strict and q>c. Now let a=q-c and b=p-c and note that n+c^2 = pq = (a+c)(b+c)=ab+ac+bc+c^2. This implies that n=ab+bc+ca where a,b,c>0 since c>0, q>c and p>c. Now it remains to show that a, b and c are pairwise distinct. If b=c, then p=2c which contradicts the fact that p is odd. If a=b, then p=q which implies that n=p^2 - c^2 = (p-c)(p+c). However, since p and c are both odd, 2|p-c and 2|p+c and hence 4|n which contradicts the fact that n is square-free. If a=c, then q=2c and 2pc-n=c^2 which is a contradiction since n is even and c is odd. Hence a, b and c are pairwise distinct as desired.
12.03.2012 19:48
Solution : Let 0<k<p , n=pl-k^2 , for natural l n=pl-k^2>\frac{p^2}{4} , so pl> \frac{p^2}{4}+k^2\ge pk , l>k Let a=k , b=l-k , c=p-k , easy to see that ab+bc+ca=lp-k^2=n Easy to check that a , b , c are distinct . done
06.11.2016 17:21
Mosquitall wrote: Solution : Let 0<k<p , n=pl-k^2 , for natural l n=pl-k^2>\frac{p^2}{4} , so pl> \frac{p^2}{4}+k^2\ge pk , l>k Let a=k , b=l-k , c=p-k , easy to see that ab+bc+ca=lp-k^2=n Easy to check that a , b , c are distinct . done wouldn't be the whole solution as n=pl-k^2 isn't the only possible case
11.07.2021 15:29
Here's my solution. Hopefully different and (imho) easier. The solution is based on reverse engineering. Wlog assume k<p (or else we can take k\to k\pmod p ). One can rewrite the desired form of n as c(a+b)+ab. So if there's such c, we must have a+b\mid n-ab\leftrightarrow a+b\mid n+a^2Why not a=k, b=p-k \hspace{2 mm} \text{and} \hspace{2 mm} c=\frac{n-k(p-k)}{p} ? Note that n+k^2>kp and clearly a\neq b. \bullet \hspace{2 mm} if a=c, that is 2kp=n+k^2\to 4\mid n, absurd. \bullet \hspace{2 mm} if b=c, that is n=p^2-k^2\to 4\mid n, absurd. So the given construction works and we're done. \blacksquare Remark. I wonder if All-Russian 2021/10.7 is inspired from this or not.