Equations Proof Solvers

Question

Can Someone solve

$$x(x+1)(x+2)...(x+2020)=1$$

analitically rather than numerically and show me all steps. Also maybe with some preconditions, like only natural solutions, so that it‘s adaptable to diophantine equations, etc.

Thank you all in advance!!!

A very nice fact seems to be true, only polynomials cubic and quartic being exceptions: all the polynomials $x(x+1)(x+2)(x+3)\cdots (x+n)=1$ have its $n+1$ roots all real. ¿Is this true for $n\ne 2,3$? — Piquito, Jan 21 '21 at 23:06
@Piquito I have added a partial answer. Feel free to give it a try and complete it. — Raffaele, Jan 22 '21 at 10:05

score 1 · Accepted Answer · answered Jan 22 '21 at 10:04

Write the equation like $$(x+1) (x+2) (x+3) (x+4) (x+5)\ldots(x+k)=\frac{1}{x}$$ The intersection between the polynomial $y=(x+1) (x+2) (x+3) (x+4) (x+5)\ldots(x+k)$ and the hyperbola $y=1/x$ are exactly $k$ if $k>3$.

This happens because $P_k(x)=(x+1) (x+2) (x+3) (x+4) (x+5)\ldots(x+k)$ has $k$ zeros and in any interval $(j-1,j);\;j\in\mathbb{Z},j<0$ there is a minimum $\xi_j$ and a maximum $\zeta_{j-1}$ for $j=\ldots,-3,-2,-1$.

TO DO

We must prove that the local minima $\xi_j$ have negative values $P(\xi_j)<\frac{1}{\xi_j}$ for $|j|>3$ therefore the curve $y=P_j(x)$ intersects the hyperbola in $|j|$ points for $|j|>3$.

$$...$$

So it is true. I am too busy in other thing. I hope yourself could finish. Good luck. — Piquito, Jan 22 '21 at 11:52

Equations Proof Solvers

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