MHB Prove a system has no real solutions

[anemone]

Prove that for any natural number $n$, the equation $y(y+1)(y+2)\cdots(y+2n-1)+(y+2n+1)(y+2n+2)\cdots(y+4n)=0$ does not have real solutions.

 

[kaliprasad]

Spoiler

Let x = y + 2n
so we get
$T_1+T_2$ where
$T_1 = (x+1)(x+2)\cdots(x+2n)$
$T_2 = (x-1)(x-2)\cdots(x-2n)$
we need to check various ranges
x = 0 then both $T_1$ and $T_2$ both are positive so sum is positive
$x \gt 2n$ or $x \lt - 2n$ Both are positive so sum is positive
$x \gt 0$ and $x\lt 2n$ each term of $T_1$ is numerically greater than corresponding term of $T_2$ ( to take care of -ve terms) so $|T_1|$ greater than $T_2$ and hence the sum > 0 (note $T_1$ is positive and $T_2$ can be -ve)
$x \gt 0$ and $x\lt 2n$ each term of $|T_2|$ s numerically greater than( this is -ve so absolute value) corresponding term of $|T_1|$ ( to take care of -ve terms) so $|T_2|$ greater than $|T_1|$ and hence the sum > 0 (note $T_2$ is positive and $T_1$ can be -ve)

we have shown that value is positive for all x so for all y and hence no solution

 

[anemone]

Very well done, kaliprasad!

On the other hand, one could also see that

Spoiler

With $x = y + 2n$,

$\begin{align*}T&=y(y+1)(y+2)\cdots(y+2n-1)+(y+2n+1)(y+2n+2)\cdots(y+4n)\\&=T_1+T_2\\&=(x+1)(x+2)\cdots(x+2n)+(x-1)(x-2)\cdots(x-2n)\end{align*}$

Note that when we expand the products of the two terms on the right, the terms of the odd powers of $x$ all cancel out, and we get a sum involving only even powers of $x$ with positive coefficients.

Hence, $T>0$ for all real $x$ and therefore there is no real solutions for the original equation.