# Proof

• September 16th 2010, 07:20 PM
kiddopop
Proof
Decide whether the following statement is true or false. If it is true, show that it is true, making your argument as general as possible. If it is false, disprove it by finding a counterexample.

For all integers a and b, if a is congruent to b (mod 35), then a^2 is congruent to b^2 (mod 7).
• September 16th 2010, 07:30 PM
undefined
Quote:

Originally Posted by kiddopop
Decide whether the following statement is true or false. If it is true, show that it is true, making your argument as general as possible. If it is false, disprove it by finding a counterexample.

For all integers a and b, if a is congruent to b (mod 35), then a^2 is congruent to b^2 (mod 7).

$a\equiv b\pmod{35}\implies a\equiv b\pmod{7}$. You can use the definition of congruence to prove that. So...
• September 16th 2010, 08:02 PM
Soroban
Quote:

Decide whether the following statement is true or false.
If it is true, show that it is true, making your argument as general as possible.
If it is false, disprove it by finding a counterexample.

$\text{For all integers }a\text{ and }b.$
. . $\text{if }a \equiv b\text{ (mod 35)},\text{ then }a^2 \equiv b^2 \text{ (mod 7)}$

Given: . $a \:\equiv\: b\text{ (mod 35)}$

This means: . $a - b \:=\:35n\:\text{ for some integer }n$

. . So we have: . $a \:=\:b + 35n$

Square both sides: . $a^2 \:=\:b^2 + 70bn + 1225n^2$

. . And we have:. . $a^2-b^2 \:=\:7(10bn + 175n^2)$ . . . a multiple of 7

Therefore: . $a^2\:\equiv\:b^2\text{ (mod 7)}$

• September 16th 2010, 08:04 PM
undefined
Well since Soroban gave a full solution, I'll give one too.

$a\equiv b\pmod{35}\implies 35|a-b\implies7|a-b\implies a\equiv b\pmod{7}\implies a^2\equiv b^2\pmod{7}$.