a.
By Fermat's little theorem, we have
5 and 7 are both prime, so and
. By Euler's theorem, we get
Now we can use the Chinese remainder theorem to solve for . Start with
- Taken mod 5, the second term vanishes and
. Multiply by the inverse of 4 mod 5 (4), then by 2.
- Taken mod 7, the first term vanishes and
. Multiply by the inverse of 2 mod 7 (4), then by 6.
b.
We have , so by Euler's theorem,
Now, raising both sides of the original congruence to the power of 6 gives
Then multiplying both sides by gives
so that is the inverse of 25 mod 64. To find this inverse, solve for
in
. Using the Euclidean algorithm, we have
64 = 2*25 + 14
25 = 1*14 + 11
14 = 1*11 + 3
11 = 3*3 + 2
3 = 1*2 + 1
=> 1 = 9*64 - 23*25
so that .
So we know
Squaring both sides of this gives
and multiplying both sides by tells us
Use the Euclidean algorithm to solve for .
64 = 3*17 + 13
17 = 1*13 + 4
13 = 3*4 + 1
=> 1 = 4*64 - 15*17
so that , and so