Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How many positive integers $(a, b, c)$ are there such that $c(a+b)^2 = a(b+c) + b(c+a)$?

I tried to play with this equation, and what I got is $c(a+b)(a+b-1)= 2ab$. However $a,b,c$ need not be primes so this does not help with this problem.

Note:- The answer is 1, but I want to approach mathematically.

share|cite|improve this question
Your rewritten factored form is good in suggesting an inequality approach. +1 for showing a start for the question. – coffeemath Apr 22 '13 at 3:52

Define $p(a,b,c)=c(a+b)^2-a(b+c)-b(c+a)$, so that you seek positive $a,b,c$ for which $p(a,b,c)=0$. There is the solution $(1,1,1)$, but not others. First note that $$p(a,b,1)=a(a-1)+b(b-1)$$ which is positive unless $a=b=1$ (going with our solution $(1,1,1)$.

Now write $p(a,b,c)$ in the form $$ca^2+a[2bc-2b-c]+b(b-1)c.$$ Here the first term is positive, and the third is nonnegative, so that provided we show that $c \ge 2$ leads to the term in square brackets being nonnegative, we can conclude $p(a,b,c)>0$ whenever $c \ge 2$.

So to finish, we have when $c$ is at least 2 that $$2bc-2b-c=c(2b-1)-2b \\ \ge 2(2b-1)-2b=2(b-1)\ge 0.$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.