Question

Which of the following equations have solution? justify your answers

a) x² =3 (mod 13⁷ ) this mean x to the 2nd power congruence to 3 mod 13 to the 7 power

b) x³ =4 (mod 11⁵.23⁹) this mean x to 3rd power congruence to 4 mod 11 to the 5 th power times 23 to the 9th power

c) x⁷=2(mod49) this mean x to the 7th power congruence to 2 mod 7 to the 2nd power

d) x⁷ = 18 (mod 49) this mean x to the 7th power congruence to 18 mod 7 to the 2nd power

Answer 1

a) 2 = 3 (mod 137) Consider x = 3 (mod 13). Write f(x)= 2²-3. So, we are looking for solutions to f(x) = 0 (mod 13). We have f (4) = 0 (mod 13). Now, f (2) = 2x and & (4)=8 70 (mod 13). By Hensel's lemma f(x) = 0 (mod 13) has a solution.

1 *54mod 119. 23°). Write f(x) = x - 4. Cost We examine f(x) = 0 (mod 115) and f(x) = 0 (mod 28?). Note that f(3) = 0 (mod 23). Now, f'(x) = 3x and & f' (3)=27=4 #0 (mod 23). Als 5 On the other hand, f (5)= 0 (mod 11) and fl (5) = 75=8960 (mod 11). By Hensel's lemma, both I f(x) = 0 (modal and f(x) = 0 (mod 239) have a solution each 1 Now, by Chinese Remainder Theorem f(x) = 0 (mod 11 6 223) has a solution.

c) x²= 2 (mod 49). By Fermat's little theorem, x= x (mod 7) for any x. Consider f(x) = -2. Now f(x) = 0 (mod 7 ) a=2 (mod 7). Now, f'(x) = 7x and f'(2) = 7.26 = 7 = 0 (mod 7). Again, by Euter's theam shem, 2 = 1 (mod 7). So, we cannot apply Hensel's temme Now, f(2)= 27-2= 126 = 28 & 0 (mod 49). Thus, there is no solution to f(x)= 0 (mod 49).

d) 2 = 18 (mod 49). - Write f(x) = x1-18. Now, f(x)=0 (mody) () 2 = 4 (modr). Again, f (x) = Te and f' (A) = 7= 0 (mod 7). So, we cannot apply Hensel's lemma. I But f (4) = 47-18=0 (mod 49). Hence, there is a solution to f(x) = 0 (mod 49).