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Let us assume the Riemann Hypothesis. DudekCarneiro, GreniéMilinovich, and MolteniSoundararajan proved in 2015proved in 2017 that for any $x\geq 2$$x>4$, there exists a prime in $(x-y,x+y)$, where $y:=\frac{1}{2}\sqrt{x}\log x+2\sqrt{x}$$[x,x+\frac{22}{25}\sqrt{x}\log x]$. Using this result, it is straightforward to prove the OP's conjecture for $x\geq 10^{40}$$x\geq 10^{47}$ (under the Riemann Hypothesis).

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Let us assume the Riemann Hypothesis. Dudek, Grenié, and Molteni proved in 2015 that for any $x\geq 2$, there exists a prime in $(x-y,x+y)$, where $y:=\frac{1}{2}\sqrt{x}\log x+2\sqrt{x}$. Using this result, it is straightforward to prove the OP's conjecture for $x\geq 10^{40}$ (under the Riemann Hypothesis).

See also my response to this related MO question.

Let us assume the Riemann Hypothesis. Carneiro, Milinovich, and Soundararajan proved in 2017 that for any $x>4$, there exists a prime in $[x,x+\frac{22}{25}\sqrt{x}\log x]$. Using this result, it is straightforward to prove the OP's conjecture for $x\geq 10^{47}$ (under the Riemann Hypothesis).

See also my response to this related MO question.

Source Link
GH from MO
  • 105.3k
  • 8
  • 293
  • 398

Let us assume the Riemann Hypothesis. Dudek, Grenié, and Molteni proved in 2015 that for any $x\geq 2$, there exists a prime in $(x-y,x+y)$, where $y:=\frac{1}{2}\sqrt{x}\log x+2\sqrt{x}$. Using this result, it is straightforward to prove the OP's conjecture for $x\geq 10^{40}$ (under the Riemann Hypothesis).

See also my response to this related MO question.