7 x^25 is to be 10 mod 83.
Fortunately 12 * 7 = 84, which is 1 mod 83. So we need x^25 to be congruent to 12 * 10 = 120 mod 83; that is, congruent to 37.
Unfortunately I don't know a way of finding a number whose 25th power is congruent to 37 mod 83, except by trial and error, which I wouldn't want to do without a program. Perhaps a programmable calculator (TI-83 for example) program, or use of the table feature, would be acceptable.
The trick to avoid the need for extra precision is to raise any trial number to the 5th power, reduce it mod 83, and then raise that number to the 5th power and again reduce mod 83. If the result is 37, you've got x as the number you started with.
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Posted by Charlie
on 2006-12-13 08:44:48 |
thoughts
