Krdytkyu

Introduction

The following argument has been advanced by one of my friends which attempts to prove the Legendre's Conjecture. I could find no flaw in the argument and so I am posting it here in the hope that people here will be able to find a flaw in the argument.

Theorem. For all sufficiently large $xinmathbb{R}$ we have, $$pi((x+1)^2)-pi(x^2)>0$$where $pi(x)$ denotes the number of primes less than or equal to $x$.

Proof.
The proof assumes the following facts,

(1) $pi(x)>dfrac{x}{ln x}$ for all $xge 17$. (see this paper for details)

(2) For all $varepsilon>0$ $dfrac{x}{ln x-(1-varepsilon)}<pi(x)<dfrac{x}{ln x-(1+varepsilon)}$ for all $xge maxleft(227,exp(frac{1.51}{varepsilon})right)$. (see here)

We will show that $pi((x+1)^2)-pi(x^2)>0$ for all sufficiently large real $x$. For this observe that, for all sufficiently large $xinmathbb{R}$ we have, begin{align}pi(x^2)<dfrac{x^2}{2(ln x-2)}end{align} (which follows from (2) by chosing $varepsilon=3$) and $$pi((x+1)^2)>dfrac{(x+1)^2}{2ln (x+1)}$$by (1). So we have, begin{align}pi((x+1)^2)-pi(x^2)&>dfrac{(x+1)^2}{2ln (x+1)}-dfrac{x^2}{2(ln x-2)}end{align}
To prove the conjecture it suffices to show that,
$$dfrac{(x+1)^2}{2ln (x+1)}>dfrac{x^2}{2(ln x-2)}$$ To prove which it suffices to show that, $$dfrac{(x+1)^2}{2(ln (x+1)-2)}dfrac{2ln(x+1)-4}{ln (x+1)}>dfrac{x^2}{2(ln x-2)}$$Which holds for all sufficiently large $x$. So we are done.

Question

What is(are) the flaw(s) in the proof? I am sure that there must be some because if the argument were this easy, it would have been proved much earlier.

Well, from the second last line to the last line you go from

$$frac{(x+1)^2}{2 ln(x+1)} >^{?} frac{x^2}{2 (ln(x) - 1)} $$

and then you multiply the numerator and denominator of the LHS by $2 (ln(x+1) - 2) = 2 ln(x+1) - 4$ to get

$$frac{(x+1)^2}{2 (ln(x+1) - 2)} frac{2 ln(x+1) - 4}{2 ln(x+1)} >^{?} frac{x^2}{2 (ln(x) - 1)} $$

and then, for some unknown reason, you multiply the LHS by a factor of $2$ to get

$$frac{(x+1)^2}{2 (ln(x+1) - 2)} frac{2 ln(x+1) - 4}{ ln(x+1)} > frac{x^2}{2 (ln(x) - 1)} $$

So yeah, the last inequality holds (for large $x$) but not the previous two inequalities, because you left out the factor of $2$ in $2 ln(x+1)$. In fact, the first two (equivalent) inequalities just completely fail, so all this argument proves is the slightly less impressive inequality

$$pi((x+1)^2) - pi(x^2) ge text{some negative number}$$

which it turns out in fact can be deduced via more direct methods.