Mathematics Asked by Mardia on November 6, 2020

The question is "Show that the solution of the equation $x^5-2x^3-3=0$ are all less than 2."

I have attempted to answer this question using proof by contradiction and I think my answer is either wrong or not a well written solution. I would like to know if I solved it right and if I did, I would like some advice on how I can improve writing proofs.

**My attempt:**

Assume to the contrary that solution of this equation is greater then or equal to 2.

Let $x=frac 2p$.

Then we have

$ (frac 2p)^5-2(frac 2p)^3-3=0 $

$ frac {2^5}{p^5} – frac {2^4}{p^3} – 3 = 0$

We now consider two case: when $p=1$ and $p<1$.

when $p=1$:

$ 2^5 – 2^4 – 3 = 32 – 16 – 3 = 13$.

Since $x = 2$ is not a solution this is a contradiction.

When $p<1$:

If $p<1$, then we know $1/p>1.$ This implies $frac {1}{p^{n+1}} > frac {1}{p^n}.$

Since $frac 1{p^3}(2^5-2^4)-3 > 2^5-2^4-3 = 13 > 0$, it is clear that the inequality

$ frac 1{p^5}(2^5)- frac 1{p^3} 2^4-3 > frac 1{p^3}(2^5-2^4)-3 > 2^5-2^4-3 = 13 > 0$ holds

Since for all $x > 2$ is not a solution this is a contradiction.

Thus, solution $x$ is less then 2.

Let $x$ be a root and $x>2$.

Thus, $$x^5-2x^3-3=x^5-2x^4+2x^4-4x^3+2x^3-3>0,$$ which is a contradiction.

Correct answer by Michael Rozenberg on November 6, 2020

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