Proof help

**lemondonut** · September 21st 2011, 06:16 PM

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

**chisigma** · September 21st 2011, 11:50 PM

Originally Posted by lemondonut

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 $a^{2} \equiv b^{2}\ \text{mod}\ 20$ , then there is a constant k for which is...

$(a+b)\ (a-b)= 20\ k$ (1)

From (1) is we derive that must be $5|(a+b)$ or $5|(a-b)$ . If $5|(a-b)$ then is $a \equiv b\ \text{mod}\ 5$ . If $5|(a+b)$ then for (1) it must be...

$(a-b)\ \frac{a+b}{5}= 4\ k \implies a-b= \frac{5}{a+b}\ 4\ k$ (2)

... so that $5|(a-b)$ ...

Kind regards

$\chi$ $\sigma$

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