Prove $ax \equiv1 \textrm{ mod } 10^k \implies aN(x) \equiv1 \textrm{ mod } 10^{2k}$

Question

For every $a \in Z$, consider the function $N:\mathbb{Z} \rightarrow \mathbb{Z}$: $$N(x)=x(2-ax)$$ Prove for every $k \in \mathbb{N}$ that $$ax \equiv1 \textrm{ mod } 10^k \implies aN(x) \equiv1 \textrm{ mod } 10^{2k}$$

Thoughts so far: We can make the following computation and find that $$aN(x)-1 = ax(2-ax)-1=-(ax-1)^2$$ which shows that $aN(x) \equiv1 \textrm{ mod } 10^k$, right? But where does $10^{2k}$ enter the picture?

@leoli1 You have the divisibilities reversed. $a\mid b$ means $a$ divides $b\ \ $ — Bill Dubuque, Jan 11 '21 at 14:59
@BillDubuque Of course, you are right (However the idea stays the same) — leoli1, Jan 11 '21 at 15:00
By the linked product rule $, A\mid B\Rightarrow A^2\mid B^2.,$ Put $,A = 10^k,\ B = ax-1\ \ $ — Bill Dubuque, Jan 11 '21 at 15:11

score 2 · Accepted Answer · answered Jan 11 '21 at 12:31

2

You're almost there:

$$ax-1=10^k q \implies (ax-1)^2=10^{2k} q^2 \equiv 0 \bmod 10^{2k}$$

answered Jan 11 '21 at 12:31

lhf

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Prove $ax \equiv1 \textrm{ mod } 10^k \implies aN(x) \equiv1 \textrm{ mod } 10^{2k}$

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