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Sep 30 at 14:25 comment added user967210 I added a new condition, but I don't know how to prove it
Sep 30 at 14:25 history edited user967210 CC BY-SA 4.0
Added a new condition
Sep 30 at 11:53 comment added user967210 @TitoPiezasIII Sorry, I had to add that further condition, as OscarLanzi explained
Sep 28 at 13:43 history edited Oscar Lanzi CC BY-SA 4.0
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Sep 28 at 9:59 comment added Oscar Lanzi I looked at the reference and it adds the condition that $2^{(p−1)/4}\equiv−1\bmod p$. Thus $2^{10}\equiv−1\bmod41$ but $2^{18}\equiv+1\bmod73$, and solvability of the negative Pell equation follows suit.
Sep 28 at 9:31 history edited Oscar Lanzi CC BY-SA 4.0
Correction made.
Sep 28 at 5:03 comment added Tito Piezas III @user967210 I just noticed that regarding prime $p \equiv 9\;\rm mod\;16$ for $x^2-2py^2=-1$, while $p=41$ is solvable, $p=16\cdot4+9=73$ and $p=16\cdot5+9=89$ are not, contrary to your last sentence. Care to explain?
Jun 22 at 18:57 comment added Will Jagy It appears Whitford skips over that one. However, on page 80 he lists conditions 1,2,3,4, but the refers to conditions 3,4,5 . Perhaps he meant to include a condition 5 with the simple $(p,q) = -1,$ or meant that as condition 4 and the one with quartic residues moved to condition 5.
Jun 22 at 18:36 comment added Will Jagy Note that Mordell's proof can be easily adapted to this: primes $p,q \equiv 1 \pmod 4$ but mutual quadratic non-residues, that is Legendre symbol $(p,q) = -1$ then there are integer solutions to $x^2 - pq y^2 = -1$ The first two semiprimes where impossibility is a surprise are $205 = 5 \cdot 41$ and $221 = 13 \cdot 17$ both times mutual residues.
Mar 25, 2022 at 17:15 history edited Glorfindel CC BY-SA 4.0
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Mar 25, 2022 at 17:15
S Mar 25, 2022 at 16:31 history edited user967210 CC BY-SA 4.0
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Mar 25, 2022 at 15:15 review Late answers
Mar 25, 2022 at 15:24
S Mar 25, 2022 at 14:54 review First answers
Mar 25, 2022 at 15:24
S Mar 25, 2022 at 14:54 history answered user967210 CC BY-SA 4.0