Timeline for Is the diophantine equation $3x^2+1=py^2$ always solvable for each prime $p\equiv 13\pmod{24}$?
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| Oct 24, 2021 at 13:39 | comment | added | GH from MO | Now assume that $t$ is odd. Then $(t-1)/2$ and $(t+1)/2$ are coprime, and their product is $3p(u/2)^2$. This leads to four cases. If $(t-1)/2=a^2$ and $(t+1)/2=3pb^2$, then $a^2-3pb^2=-1$, so $(-1/3)=1$, contradiction. If $(t-1)/2=3a^2$ and $(t+1)/2=pb^2$, then $pb^2-3a^2=1$, which is the conclusion we want. If $(t-1)/2=pa^2$ and $(t+1)/2=3b^2$, then $pa^2-3b^2=-1$, so $(-1/3)=1$, contradiction. If $(t-1)/2=3pa^2$ and $(t+1)/2=b^2$, then $b^2-3pa^2=1$, which contradicts the minimality of $(t,u)$. | |
| Oct 24, 2021 at 13:30 | comment | added | GH from MO | Nice proof! Let me add more detail. First assume that $t$ is even. Then $t-1$ and $t+1$ are coprime, and their product is $3pu^2$. This leads to four cases. If $t-1=a^2$ and $t+1=3pb^2$, then $a^2-3pb^2=-2$, so $(-2/p)=1$, contradiction. If $t-1=3a^2$ and $t+1=pb^2$, then $pb^2-3a^2=2$, so $(2/3)=1$, contradiction. If $t-1=pa^2$ and $t+1=3b^2$, then $3b^2-pa^2=2$, so $(6/p)=1$, contradiction. If $t-1=3pa^2$ and $t+1=b^2$, then $b^2-3pa^2=2$, so $(2/3)=1$, contradiction. I continue in the next remark. | |
| Oct 24, 2021 at 11:57 | history | edited | GH from MO | CC BY-SA 4.0 |
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| Oct 24, 2021 at 11:56 | comment | added | Franz Lemmermeyer | @GH: thanks - corrected. | |
| Oct 24, 2021 at 11:55 | history | edited | Franz Lemmermeyer | CC BY-SA 4.0 |
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| Oct 24, 2021 at 11:39 | comment | added | GH from MO | Your $\eta$ does not lie in $\mathbb{Q}(\sqrt{3p})$. It lies in the biquadratic field $\mathbb{Q}(\sqrt{3},\sqrt{p})$. | |
| Oct 24, 2021 at 10:07 | history | answered | Franz Lemmermeyer | CC BY-SA 4.0 |