If $a ≡ b$ (mod $n$), prove that $a^5 ≡ b^5$ (mod $n$)

Question

If $a ≡ b$ (mod $n$), prove that $a^5 ≡ b^5$ (mod $n$).

I know there is some way of solving this using binomial expansion but it has me stumped.

Answer 1

Note if $a\equiv b\bmod n$ then $n$ divides $a-b$ thus there exists an integer $k$ satisfying $(a- b)=nk$ therefore $nk(a^4+a^3b+a^2b^2+ab^3+b^4)=a^5-b^5$ which means $a^5-b^5$ is a multiple of $n$ i.e. we have $a^5\equiv b^5\bmod n$.

If you are just learning modular arithmetic it might be more helpful to try and prove that whenever $u\equiv v\bmod n$ and $x\equiv y\bmod n$ then $ux\equiv vy\bmod n$ this would answer your question as well.

Answer 2

$a ≡ b $ (mod n)

$a^5-b^5 = (a-b)(a^4+a^3b+a^2b^2+ab^3+b^4)≡0 $ (mod n)

$a^5≡b^5 $ (mod n)

Answer 3

Notice that, for all $a,b,k \in \mathbb{R}$, $$a^k−b^k=(a−b)(a^{k−1}+a^{k−2}b+…+ab{k−2}+b{k−1}).$$ (That can be proven using induction) Now if $a≡b$ (mod n), then we have that $a=b+kn$, for some $k∈\mathbb{Z}$, by the Theorem of Division with Remainder, and hence we have that $a−b=kn$. Therefore we have that $$n∣(a−b)$$ and of course, that means that we can multiply the right-hand side by any number and still the statement would be true. As such, $$n∣(a−b)(a^{k−1}+a{^k−2}b+…+ab^{k−2}+b^{k−1})=a^k−b^k.$$ Therefore, we have that $$rn=a^k−b^k$$ for some $r∈\mathbb{Z}$ and hence $$a^k≡b^k (mod \ n).$$

Let $k$ be equal to $5$ and you'd obtain the answer.

Answer 4

Prove that if $a \equiv b\pmod n$ and $c\equiv d\pmod n$ then $ac \equiv bd \pmod n$.[1]

Once you know that then $a\equiv b \pmod n \implies a*a=a^2\equiv b*b \equiv b^2 \pmod n\implies a*a^2\equiv b*b^2 \pmod n \implies ...... a*a^k=a^{k+1}\equiv b*b^k \equiv b^{k+1}\pmod n$.

....

But to use binomial th, which you don't need.

$a \equiv b\pmod n$ meand $b = a+ mn$ for some intger $m$.

So $b^k = (a + mn)^k = a^k + ka^{k-1}mn + {n\choose 2}a^{k-2}m^2n^2 + ...... + {n\choose n}a^2m^{k-2}n^{k-2} + kam^{k-1}n^{k-1} + m^kn^k=$

$b^k + a^k + [ka^{k-1}m + {n\choose 2}a^{k-2}m^2n + ...... + {n\choose n}a^2m^{k-2}n^{k-3} + kam^{k-1}n^{k-2} + m^kn^{k-1}]\equiv a^k \pmod n$.

.....

And if you want to be slick.... (this what the cool kids are all doing).....

$a^k - b^k = (a-b)(a^{k-1} + a^{k-2}b + .... ab^{k-2}+ b^{k-1})$

So $a\equiv b \pmod n \implies n|(a-b) \implies n| (a-b)(a^{k-1} + a^{k-2}b + .... ab^{k-2}+ b^{k-1})\implies n|a^k - b^k \implies a^k \equiv b^k \pmod n$.

====

[1] Pf: $a\equiv b\pmod n$ if there is an $m$ so that $a=b + mn$ and $c\equiv d \pmod n$ if there is a $k$ so that $c =d + kn$.

So $bd = (b+mn)(d+kn) = bd + dmn + bkn + mkn^2 = bd + n(dm+bk +mkn)$

So $bd \equiv ac\pmod n$.

=====

ANd just to be specific.

$a \equiv b \pmod n$ means that

$n|b-a$ means that $\frac {b-a}n = m$ for some integer $m$.

$b-a = mn$ and $b = a + mn$.

So $b^5 = (a+mn)^5 = a^5 + 5a^4mn + 10a^3m^2n^2 + 10a^2m^3n^3 + 5am^4n^4 + m^5n^5$

$= a^5 + n[5a^4m + 10a^3m^2n + 10a^2m^3n^2 + 5am^4n^3 + m^5n^4]$

so

$b^5-a^5 = n[5a^4m + 10a^3m^2n + 10a^2m^3n^2 + 5am^4n^3 + m^5n^4]$

so

$\frac {b^5-a^5}n = [5a^4m + 10a^3m^2n + 10a^2m^3n^2 + 5am^4n^3 + m^5n^4]$ which is an integer

So $n|b^5-a^5$.

Which is what the definition of $a^5 \equiv b^5 \pmod n$ is.

(Everyone really should work a complete definition through a few times in their lives.

Answer 5

It is conceptually simpler to prove by strong induction on $k$ that $$\forall k\ge 1,\quad a \equiv b\pmod n \implies a^k\equiv b^k\pmod n $$

• For $k=0, 1$, there's nothing to prove.
• Suppose the implication is true up to some $k\ge 1$. Then if $a\equiv b\bmod n$, we know that $a^k\equiv b^k\bmod n$, i.e. $a^k-b^k$ is divisible by $n$. Now , $$a^{k+1}-b^{k+1}=(a+b)(a^k-b^k)-ba^k+ab^k=(a+b)(a^k-b^k)-ab(a^{k-1}-b^{k-1})$$ is divisible by $n$ since $a^k-b^k$ and $a^{k-1}-b^{k-1}$ are by the (strong) inductive hypothesis.