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The conjecture of Goldbach (1752)

"Every odd integer can be written in the form $p+2a^2$ where p is a prime or 1 and a is a natural number (can be even 0)."

has the smallest exception $5777$. While not allowing $p=1$ the formula give the following exceptions less than $10,000,000$:

1 5777 5993 4946453 5293163 5651423 6017447 6098663 6392393 6420563
6429467 7071803 7072865 7081223 7103447 7217417 7277753 7341677
7499447 7532687 7571537 7642637 7727657 7809383 7825643 7844183
7860815 7935173 7970267 8063393 8111033 8226863 8286527 8344247
8409437 8464367 8498393 8542397 8552147 8562857 8654093 8725973 
8772227 8848883 8851403 8863703 8865593 8867663 8869907 8885675
8952797 8957747 8971775 8985503 9166637 9200573 9233333 9309857
9325103 9341147 9409787 9447533 9470147 9481763 9492203 9519269
9534653 9552155 9553367 9564347 9707303 9760307 9843275 9882023
9897017 9911393 9916337 9948617 9958307

It's trivially true that all odd numbers greater than $1$ can be written in the form $p+2a$, $p$ prime, and I therefore conjecture:

There is a real number $\gamma>1$ such that every odd number greater than $1$ can be written in the form $p+2\lfloor a^\gamma\rfloor$. Where $p\in\mathbb P$ and $a\in\mathbb N$

Is this possible to prove?

For $\gamma=1.5\;$ I found no exceptions less than $100,000,000$.

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    $\begingroup$ You can look up "Piatetski-Shapiro primes" to see some techniques used to analyze a sequence of integers like $\lfloor a^x \rfloor$. $\endgroup$ – Greg Martin Nov 4 '16 at 18:40
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Of course this is true. $a^\gamma $ has an image of $\mathbb {R}^+$. Thus for any $n\in\mathbb{N}$ we have that there will be a solution to $1/2 (n-p)=a^\gamma$ for $\gamma>1\in\mathbb {R} $ so long as we choose $p $ such that $n-p>2a $, and it follows that $n=p+2\lfloor a^\gamma\rfloor $ will have a solution.

EDIT
Answer is obsolete with new question edit.

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