Timeline for Diophantine equation with no integer solutions, but with solutions modulo every integer
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Apr 13 at 17:45 | history | edited | Alexey Ustinov | CC BY-SA 4.0 | added 6 characters in body |
| Feb 3, 2019 at 5:54 | comment | added | KConrad | A simpler example than $y^2 = x^3 - 51$ of such an equation with no integral solution and with a rational solution is $y^2 = x^3 + 11$, with rational solution $(-7/4,19/8)$. That it has no integral solution can be proved with just congruences and knowing when $-1 \equiv \Box \bmod p$, no need for class numbers. See Theorem 2.3 of kconrad.math.uconn.edu/blurbs/gradnumthy/mordelleqn1.pdf. The rational solution shows there is a solution mod $p^r$ for odd primes $p$, and to get a solution mod powers of $2$ there is a $2$-adic solution $(x,2)$ where $x^3 = -7$ with $x \equiv 1 \bmod 2$. | |
| Feb 3, 2019 at 2:31 | comment | added | KConrad | The point of my previous comment was to have a rational point and no integral point, but the rational point doesn't matter. For many nonzero integers $k$ (e.g., $k = \pm 6$) the equation $y^2 = x^3 + k$ has no integral solutions, but for every nonzero integer $k$, the congruence $y^2 \equiv x^3 + k \bmod m$ has a solution for all $m \geq 2$. It suffices by Chinese remainder theorem to show there is a solution when $m$ runs through prime powers, and my answer at mathoverflow.net/questions/134352 does that. (At math.stackexchange.com/questions/875983 I show solns mod $p$.) | |
| Nov 28, 2010 at 7:19 | comment | added | KConrad | A Weierstrass equation in this spirit is y^2 = x^3 - 51 (which BCnrd told me about, and I think he got it from Venkatesh). It has the rational solution (1375/9,50986/27) and it can be solved mod 3^r for any r using x = 1 and for y a square root of -50 mod 3^r (which exists since -50 = 1 mod 3). Thus as a congruence this equation has a solution mod m for any m. That there is no Z-solution follows from Q(sqrt(-51)) having class number 2, which is rel. prime to 3, by the same kind of method used to find the Z-solutions to y^2 = x^3-2. | |
| Nov 28, 2010 at 0:45 | comment | added | Kevin Buzzard | Yours seems to me like the "best" answer so far though (simplest equation, smallest coefficients). | |
| Nov 28, 2010 at 0:45 | comment | added | Kevin Buzzard | KConrad: it would have been shorter to say "$(1/3,1/3)$ and $(3/5,1/5)$" than "Look for a rational solution....both cases)." :-) | |
| Nov 27, 2010 at 19:34 | history | answered | KConrad | CC BY-SA 2.5 |