Proving Infinitely Many Pairs of Positive Integers for Sum Equation

[sennyk]

I have it. Suppose

$$m^2 - (Zn - 1)m + n(n+1) = 0$$

This has two roots

$$m^2 - (m_1 + m_2) + m_1m_2 = 0$$

I'm not seeing how you make this leap. Please explain.

 

[Ben Niehoff]

I'm not seeing how you make this leap. Please explain.

Whoops, there's a typo. It should be:

$$m^2 - (m_1 + m_2)m + m_1m_2 = 0$$

It's directly from the fundamental theorem of algebra. If a second-degree polynomial has roots r_1 and r_2, then

$$\begin{array}{rcl}(x - r_1)(x - r_2) & = & 0 \\ x^2 - r_1x - r_2x + r_1r_2 &=& 0 \\ x^2 - (r_1 + r_2)x + r_1r_2 &=& 0\end{array}$$

 

[robert Ihnot]

sennyk, while he did make a typo, you have got to look at symmetric functions as they apply to the roots of a polynominal.

Take the equation X^3-1 = 0. This equation has three roots x=1, $$X=\frac{-1\pm\sqrt-3}{2}$$

Question: What is the sum of the three roots and what is their product?

Answer: In the equation X^3-bX^2+cX-d, the sum of the roots equals -b, and the product of the roots equals -d. Since b=0 the sum of the roots is 0 and since -d =1 the product of the roots is 1.

We get the above form from multiplying out (x-r)(x-s)(x-t), where r,s,t represent the three roots.

 

[sennyk]

The typo completely threw me off. I know how to find the roots of a polynomial. I'm not a complete amateur. :)

 

[robert Ihnot]

O.K., sorry