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In a forum post the following conjecture was proposed with no background information.

For all positive non-perfect-square number $a$, there exists integer $b$ such that $$ a^3 - b^2 \text{ is a prime number.} $$ By prime numbers, here we only take positive primes into consideration.

E.g., $6^3 - 5^2 = 191$, $8^3 - 3^2 = 503$, and $14^3 - 5^2 = 2719$ are all primes.

I have verified for $a \le 5000$ with no exception. The non-perfect-square condition is needed as $a = 4, 25$ gives no valid $b$'s, but is not exact as $9^3 - 26^2 = 53$ and $16^3 - 63^2 = 127$ are primes.

An early pattern suggests that taking $b = a - 1$ suffices in most cases, but it fails at $a = 14$ (smallest) as $14^3 - 13^2 = 2575$ is a multiple of $5$. And, of course, when $a$ is sufficiently large, this fails mostly.

I cannot prove it, so I suggest that this is an open problem. Also, in a probabilistic sense, this may be harder than the Goldbach's conjecture. (I know this is very inaccurate.)

However, I searched about it but found no useful information. So I post it here, and hopefully someone can afford some references.

Edit: Thanks to MSE's "relation finding", I found (in this question) that this is a special case of Hardy and Littlewood's "conjecture H" (Hardy and Littlewood, 1923). Though this only considers cubes. I'll leave the question as is, or wait for the mods taking actions.

PinkRabbit
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  • Regardless of that, if this is true then I seriously doubt that any technique exists to prove it. Of course, a deeper search might find a counterexample. – lulu Oct 14 '24 at 20:52
  • @lulu sorry I'm a bit dazed then, I'll correct it. – PinkRabbit Oct 14 '24 at 20:54
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    Do you allow $a^3-b^2$ to be negative primes, or only want positive primes? – Thomas Andrews Oct 14 '24 at 20:57
  • @ThomasAndrews The original problem seems to ignore negative primes. But if considering negative primes gives useful facts, I appreciate it. – PinkRabbit Oct 14 '24 at 21:02
  • Post edit, the question is clear (though I certainly supposed you meant positive primes). Beyond a (much) deeper search, though, I doubt you'll find much. Assuming it is true of course, and your search seems deep enough to preclude cheap counterexamples. – lulu Oct 14 '24 at 21:06
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    works for $a$ up to a million. Out of that, about 51000 times, the largest possible value for $b$ works, that being $\left \lfloor a^{3/2} \right \rfloor$ 999938 sqrt 999907001 999907001 prime 882947671 ////// 1000062 sqrt 1000093001 1000093001 prime 883052327 ////// 1000124 sqrt 1000186005 1000186005 prime 1532046599 – Will Jagy Oct 14 '24 at 21:39
  • Well, if you only want positive primes, there are only a finite number of $b$ to try for each $a,$ but there are infinitely many possible $b$ which makes it virtually impossible to check completely for any $a.$ – Thomas Andrews Oct 15 '24 at 02:28
  • here's a well known formula that would not produce a single prime. $n^3 + T_{n-1}^2 = T_{n}^2$ which can be rewritten $n^3 - T_{n}^2 = -T_{n-1}^2$. if negative numbers are allowed with $T_{n}$ a triangular number. – user25406 Oct 15 '24 at 12:19
  • The exceptions where $a=k^2$ works are explained as $(k^3)^2-b^2=(k^3-b)(k^3+b)$ which cannot be prime unless $k^3-b=1$ – Keith Backman Oct 15 '24 at 14:19
  • Check mod 6 maybe then translate a^3 mod 18 and b^2 mod 12 – Roddy MacPhee Apr 01 '25 at 17:45

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