I have the following question, actually from the book Rational Points on Elliptic Curves by Silverman and Tate;
Prove that for every exponent e≥1, the congruence
x2+1≡0(mod 5e)
has a solution xe∈Z/5eZ. Prove further that these solutions can be chosen to satisfy
x1≡2(mod 5),andxe+1≡xe(mod 5e)for all e≥1.
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 e≥1, the congruence x2+1≡0 (mod 5e) has a solution xe∈Z/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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