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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.

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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

1 Answer 1

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 $$2bc-2b-c=c(2b-1)-2b \\ \ge 2(2b-1)-2b=2(b-1)\ge 0.$$

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