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Let $a$ and $b$ be integers with $a ≡ b \pmod 7$. Is $a^2 ≡ b^2 \pmod 7$? Justify by giving a proof or a counterexample.

I actually have no clue how to even begin tackling this question. How would I go about it?

Alex M.
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Winston Nguyễn
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  • With a little bit of knowledge about quotient rings, or maybe just a little group theory, this becomes as simple as proving that $a=b$ implies $a^2=b^2$. But I'm guessing you're supposed to do this without using group theory. – David K Dec 17 '15 at 22:20

4 Answers4

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Definition

$$\rm x\equiv y\pmod m \iff m|x-y.$$

HINT:

$$\rm a-b=7k \rightarrow (a-b)(a+b)=a^2-b^2=(7k)q.$$

For some $\rm k,q\in \Bbb Z$.

YoTengoUnLCD
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$$a\equiv b\pmod 7$$ means, by definition, that $a-b$ is a multiple of 7.
$$a^2\equiv b^2\pmod 7$$ means that $a^2-b^2$ is a multiple of 7.

Is there some way you could show that if $a-b$ is a multiple of 7 then $a^2-b^2$ must also be a multiple of 7?

MJD
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Actually, it's quite simple. $$a^2\equiv b^2 \pmod7$$ $$a^2- b^2\equiv0 \pmod7$$ $$(a-b)(a+b)\equiv 0\pmod 7$$ Since $a-b\equiv0\pmod7$, the equivalence holds.

Or, basically, being $(a-b)$ a multiple of $7$, also $(a-b)(a+b)=a^2-b^2$ is a multiple of $7$. Conclude by definition of congruence modulo $7$.

Ottavio
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In general if $a = b \mod(n)$ then $a*c = b*c \mod(n)$ (but not necessarily vice versa).

Pf: $a = b \mod(n) \implies n|a - b \implies n|c(a - b)=ca - cb \implies ac = bc \mod (n)$.

Also $a = b \mod(n); b = c \mod(n) \implies a = c \mod(n)$.

Pf: $n|a - b; n|b - c \implies n|(a - b) + (b - c) = a - c.$

So $a = b \mod(n) \implies ab = b^2 \mod(n)$. $a = b \mod (n) \implies a^2 = ab \mod(n).$ So $a^2 = b^2 \mod(n)$.

fleablood
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