Timeline for A question regarding simultaneous congruences
Current License: CC BY-SA 3.0
3 events
| when toggle format | what | by | license | comment | |
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| Apr 9, 2012 at 19:12 | comment | added | Noam D. Elkies | Not only is it at most the number of coprime pairs in $[1,N]$ but indeed exactly that number. If $a/b \equiv c/d \bmod p$ then $ad-bc$ is a multiple of $p$, and $|ad-bc| \leq N^2 < p$ so $ad-bc=0$ and $a/b=c/d$. | |
| Apr 9, 2012 at 17:24 | vote | accept | Stanley Yao Xiao | ||
| Apr 9, 2012 at 17:19 | history | answered | Seva | CC BY-SA 3.0 |