Exercise 2.1.39 (Find $a,b$ such that $m^2 \mid a^2 -b^2, m
mid a-b, m
mid a+b$.)

Question

If $a$ and $b$ are real numbers such that $a^2 = b^2$, it is well known that $a=b$ or $a=-b$. Give an example to show that if $a^2 \equiv b^2 \pmod{m^2}$ for integers $a,b$ and $m\geq 2$, it does not necessarily follow that $a\equiv b \pmod m$ or $a \equiv -b \pmod m$.

Richard Ganaye · Accepted Answer

\begin{proof} $77^2 - 23^2 = 5400 \equiv 0 \pmod{30^2}$, but $77 
ot \equiv 23 \pmod {30}, 77 
ot \equiv -23 \pmod {30}$.
\end{proof}

Exercise 2.1.39 (Find $a,b$ such that $m^2 \mid a^2 -b^2, m \nmid a-b, m \nmid a+b$.)

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Exercise 2.1.39 (Find $a,b$ such that $m^2 \mid a^2 -b^2, m \nmid a-b, m \nmid a+b$.)

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