This topic contains 4 replies, has 2 voices, and was last updated by Yash Agarwal 3 years ago.
October 6, 2016 at 2:45 pm #351
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Hey Math Circle Students!
These math problems are for 20 points and really develop your skills as mathematicians to understand higher order thinking beyond just math competitions and AMC 8/Mathcounts, etc. These focus on observing and proving patterns and making conjectures!
If you don’t understand one of the questions, feel free to post in the questions forum here
1. First find the solutions to all of the following modular congruences:
(a) $x^2 \equiv 2 \bmod 7$
(b) $x^2 \equiv 1 \bmod 13$
(c) $x^3 \equiv 1 \bmod 7$
(d) $x^2 \equiv 1 \bmod 15$
(e) $x^2 \equiv 1 \bmod 21$
2. What do you notice about the number of solutions to the equations with prime moduli? What about the number of solutions to the equations with composite moduli? If there were no mod signs, what could you tell about the solutions?
3. Does the fundamental theorem of algebra that a $n$th degree polynomial has at most $n$ roots apply to these problems? Does it work for some, if not then which ones does it not work for and why?
(Note: The congruence $x^2 \equiv 1 \bmod 7$ is equivalent to finding the roots of $x^2-1$ in mod 7.)
If you discover a pattern or you manage to prove one of the problems, post your discoveries below! These are really open-ended so you can post any progress you have made on the problems!
October 12, 2016 at 6:39 pm #364
e)1October 18, 2016 at 5:41 am #375
Good start Prem Dhoot, you have been awarded 10 points! But, recalculate your cases b-e, find the actual numbers that are solutions to the equations.October 19, 2016 at 7:29 am #378
Wait, for #1, does x have to be an integer? It is not specified.October 19, 2016 at 7:33 am #379
The equations with prime moduli always have 2 solutions. The equations with composite moduli have 4 solutions. If there were no mod sign, that would mean:
Example: e) x^2 = 1(mod 21)
It would be:
x^2 = 1 + 21
x^2 – 1 = 21
(x+1)(x-1) = 21
So, the x^2 – 1 would just equal the divisor in mod equations.
This response is written by Yash Agarwal