Friday, 22 March 2019

number theory - Modular equations and congruences



I have the following question, actually from the book Rational Points on Elliptic Curves by Silverman and Tate;




Prove that for every exponent e1, the congruence
x2+10(mod 5e)



has a solution xeZ/5eZ. Prove further that these solutions can be chosen to satisfy



x12(mod 5),andxe+1xe(mod 5e)for all e1.




The best way to go about it is supposed to be induction on e. So I have gone through the initial step, i.e. for e=1 and shown that there is indeed a solution to the congruence in Z/5eZ, namely x=2 or x=3. However my problems begin from here on out. I just don't know how to proceed with this problem. This question is actually already (Prove that for every exponent e1, the congruence x2+10 (mod 5e) has a solution xeZ/5eZ). I just can't follow the answer at all and was hoping someone could provide a more basic answer or methodology I could follow. Thank you for your help!


Answer



Starting with x_1\equiv 2 \mod 5 and x_2\equiv x_1 \mod 5, we have x_2=5k+2 for some integer k. Then



x_2^2+1 = 25k^2 + 20k + 5 \equiv 20k+5 \mod 25



So k \equiv 1 \mod 5 and x_2 \equiv 7 \mod 25



Then wash, rinse and repeat ...




x_3\equiv x_2 \mod 25



\Rightarrow x_3 = 25k + 7



\Rightarrow x_3^2 + 1 = 625k^2 + 350k + 50 \equiv 100k + 50 \mod 125



\Rightarrow k \equiv 2 \mod 5



\Rightarrow x_3 \equiv 57 \mod 125




etc.



The general case is known as Hensel's lemma.


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