For all integers a and b, if a^2 is congruent to b^2 (mod 20), then a is congruent to b (mod 5). prove true or false
If If $\displaystyle a^{2} \equiv b^{2}\ \text{mod}\ 20$, then there is a constant k for which is...
$\displaystyle (a+b)\ (a-b)= 20\ k$ (1)
From (1) is we derive that must be $\displaystyle 5|(a+b)$ or $\displaystyle 5|(a-b)$. If $\displaystyle 5|(a-b)$ then is $\displaystyle a \equiv b\ \text{mod}\ 5$. If $\displaystyle 5|(a+b)$ then for (1) it must be...
$\displaystyle (a-b)\ \frac{a+b}{5}= 4\ k \implies a-b= \frac{5}{a+b}\ 4\ k$ (2)
... so that $\displaystyle 5|(a-b)$...
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$\displaystyle \chi$ $\displaystyle \sigma$