# How to prove that if $a\equiv b \pmod{kn}$ then $a^k\equiv b^k \pmod{k^2n}$

What I have done is this:

$a\equiv b \pmod{2n}$,

$a=b+c\times2n$, for some $c$,

$a^2=b^2+2b\times c\times2n+c^2\times2^2n^2$,

$a^2-b^2=(b\times c+c^2n)\times4n$, then

$a^2\equiv b^2\pmod{2^2n}$.

I think that this is right: what I DON’T understand is how to generalize this to:

$a\equiv b\pmod{kn}\Rightarrow a^k\equiv b^k \pmod{k^2n}$.

Please give me a hint.

## migrated from crypto.stackexchange.comDec 1 '11 at 13:30

This question came from our site for software developers, mathematicians and others interested in cryptography.

• Hi @gurghet - I've migrated your question here as it has no direct relation to cryptography as is - so this is the best place to get a good answer. – user892 Dec 1 '11 at 13:32

## 3 Answers

Since $a\equiv b\pmod{kn}$, we have $$a=b+ckn$$ for some integer $c$. Now taking the $k$th power on both sides, we have $$a^k=(b+ckn)^k.$$ By the binomial theorem, the right hand side is given by $$(b+ckn)^k=b^k+\sum_{i=1}^k{k\choose i}(ckn)^ib^{k-i}.$$ For $i\geq 2$, it is clear that ${k\choose i}(ckn)^ib^{k-i}$ is divisble by $k^2n$. On the other hand, for $i=1$, we have ${k\choose i}(ckn)^ib^{k-i}=ck^2nb^{k-1}$, which is also divisble by $k^2n$. Therefore, by the above equality, we have $$(b+ckn)^k=b^k+k^2nN$$ for some integer $N$. Combining all these, we have $$a^k=(b+ckn)^k=b^k+k^2nN,$$ that is $$a^k\equiv b^k \pmod{k^2n}.$$

• Thanks but I’m not sure about the binomial theorem. Is that right? I think it misses a $b^{k-i}$ in the sum. – gurghet Dec 1 '11 at 16:01
• It works anyway! Thanks! – gurghet Dec 1 '11 at 16:09
• @gurghet: Yes, you are right! I missed the term $b^{k-i}$. See my edited answer. Thanks! – Paul Dec 1 '11 at 21:52

Hint  It's a special case of:  a root of a polynomial is a double root if the derivative vanishes.

Thus to prove that $$\rm\ \ k\mid a-b\ \Rightarrow\ k^2 \mid a^k-b^k\ =\ (a-b)\ \dfrac{a^k-b^k}{a-b},\$$ it suffices to prove

that $$\rm\,k\,$$ divides the second factor $$\rm\,g.\,$$ Theorem $$\Rightarrow\rm\, g\equiv\, (a^k)'\equiv \color{#c00}k\, a^{k-1}\equiv\,0 \pmod{\!\color{#c00}k}.\$$ QED

Theorem $$\$$ For $$\rm R$$ a ring and $$\rm\,f(x)\in R[x]\quad\$$ [Universal Polynomial Derivative Formula]

$$\rm\begin{eqnarray}{}&\rm g(x,y)\ &=&\rm\ \frac{f(x)-f(y)}{x-y}\ \in\ R[x,y]\\[.3em] \Rightarrow\ &\rm g(x,x)\ &=&\rm\ f'(x)\ \in\ R[x] \end{eqnarray}$$

Proof $$\$$ By $$\rm\,R$$-linearity of the derivative it suffices to verify it for a monomial $$\rm\ f(x) = x^k\,.$$

$$\begin{eqnarray}{}\rm &\rm g(x,y)\ &=&\rm\ \frac{x^k-y^k}{x-y}\, =\ x^{k-1} + x^{k-2}\,y +\, \cdots\, + x\,y^{k-2} + y^{k-1}\\[.3em] \Rightarrow\ &\rm g(x,x)\ &=&\rm\ k\ x^{k-1} =\ f'(x)\quad\ {\bf QED} \end{eqnarray}\qquad$$

Remark  To elaborate on a subtlety mentioned in a comment, $$\rm\,g(x,y)\in R[x,y]\,$$ means that $$\rm \,g(x,y)\,$$ denotes a polynomial in the indeterminates $$\rm\,x,y,\,$$ whose coefficients lie in the ring $$\rm\,R\,$$ (recall that $$\rm\,x-y\,$$ divides $$\rm\,f(x)-f(y)\,$$ in $$\rm\,R[x,y]\,$$ by the Factor Theorem). Being a polynomial its value is well-defined for all values of $$\rm \,x,y\,$$ so we can "evaluate" it at $$\rm\,y = x\,$$ (as is exemplified by the concrete case $$\rm\,f(x) = x^k$$ used in the proof).

In effect we exploit properties of polynomial rings to universally cancel an "apparent singularity" before evaluating. In more complicated contexts this can lead to nontrivial simplifications, e.g. replacing more complicated topological arguments by trivial algebraic arguments, e.g. see the discussion of the purely algebraic proof of Sylvester's determinant identity linked here.

• It is worth emphasizing that $g(x,y)=\dfrac{f(x)-f(y)}{x-y}\in R[x,y]$ is NOT the imperative "take (f(x)-f(y)) and divide by (x-y)", but the polynomial in two variables (that's what "$\in R[x,y]$" means) which when multiplied by (x-y) results in f(x)-f(y) -- this is why g(x,x) is not undefined. – Vladimir Sotirov Feb 11 '12 at 17:39
• @VladimirSotirov Actually it does denote the result of that polynomial division. But evaluating it at "apparent singularities" requires that we compute the quotient before we evaluate it. I added a remark elaborating on this subtlety. – Bill Dubuque Jan 4 at 16:16

Since

$a^k-b^k=(a-b)(a^{k-1}+a^{k-2}b+\ldots b^{k-1})$

we need only prove that $a^{k-1}+a^{k-2}b+\ldots + b^{k-1}$ is divisible by $k$. But since $a \equiv b \ (\text{mod} \ k)$ we see that

$a^{k-1}+a^{k-2}b+\ldots + b^{k-1} \equiv a^{k-1}+a^{k-1}+\ldots +a^{k-1} \equiv ka^{k-1} \equiv 0 \ (\text{mod} \ k).$