# IMO 1979 problem

The question is $$\text{If }\, p, \ q\in \mathbb{N}, \;1-\frac12+\frac13-\frac14-\dotsb-\frac{1}{1318}+\frac{1}{1319}=\frac{p}{q}.\qquad \text{Prove that } 1979\mid p.$$

So my solution went like this:

$$S=1-\frac12+\frac13-\frac14-\dotsb-\frac{1}{1318}+\frac{1}{1319}\Rightarrow S=\sum_{k=1}^{1319}\frac{1}{k}-2\sum_{k=1}^{659}\frac{1}{2k}=\sum_{k=660}^{1319}\frac{1}{k}$$ Now this is equivalent to finding the sum of the roots of the equation $$\left(x-\frac{1}{660}\right)\left(x-\frac{1}{661}\right)\dots \left(x-\frac{1}{1319}\right)=0$$.

Rewriting this equation, we get $(660x-1)(661x-1)\dotsc(1319x-1)=0$ (we can discard the $\displaystyle \frac{1319!}{659!}$ since $1979$ is a prime). Calculating the sum of the roots, we get the required value to be equal to $(1320+659)\times 330$ which is obviously divisible by $1979$ (again discarding the coeffecient of $x^{660}$).

Now my question is: Using the same method of converting the problem into finding the sum of roots of a polynomial, we get from the original equation something like this: $$(x-1)\left(x+\frac{1}{2}\right)\dots \left(x-\frac{1}{1319}\right)=0 \Rightarrow (x-1)(2x+1)\dotsc(1319x-1)=0$$

So, the sum of roots would be $-1+2-3+4-\dotsc+1318-1319=-660$ But this is not divisible by $1979$. Why does the former solution work while the latter doesn't? Is there a flaw in my reasoning?

Note: I know that the question can be solved by some other methods. Please only explain my given doubts and don't instead give another solution

I realised my mistake after some hours of pondering over it.

(In the 1st equation) To get the required sum I would need to take calculate $\displaystyle -\frac{\text{coeffecient of }x^{659}\,}{\text{coeffecient of }x^{660}}$. However what I've done is calculate this: $\displaystyle -\frac{\text{ coeffecient of }x}{\text{coeffecient of }x^{660}}$ which is actually equal to $\text{sum of roots taken }659 \text{ at a time}$. It was just a coincidence that the value I calculated was divisible by $1979$.

The correct way, indeed, would be through the link supplied above

• How is the sum of the roots $(1320+659)\cdot 330$ or $-660$? – Hagen von Eitzen Jan 11 '14 at 16:01
• The first formula should have a negative sign for $1/4$, rigth? – leonbloy Jan 11 '14 at 16:04
• I believe that what user117913 doing is finding the coefficient of the second term in the polynomial, and using the reasoning that that coefficient must be the negative of the sum of the roots multiplied by the leading coefficient. So what is being calculated is the product of the sum of roots and the leading coefficient, and since the leading coefficient isn't divisible by 1979, the sum of roots must be. – Glen O Jan 11 '14 at 16:18
• @GlenO But that second term is way bigger. It is the sum of $>600$ positive integers, each of which is at least $\frac{1318!}{659!}$. – Hagen von Eitzen Jan 11 '14 at 16:24
• @HagenvonEitzen - yeah, I just realised I made a mistake. The calculation is for the second-last term, not the second term. I have to think about it a bit more. – Glen O Jan 11 '14 at 16:25