Question

Prove that

$$\begin{gathered} \mathbf{\left(}\mathit{a}\mathbf{\right)}\mathbf{}\mathbf{}{\left(n-a\right)}^{\mathbf{2}}\mathbf{=}{\mathit{a}}^{\mathbf{2}}\left(modn\right) \\ \mathbf{\left(}\mathit{b}\mathbf{\right)}\mathbf{}\mathbf{}{\left(2n-a\right)}^{\mathbf{2}}\mathbf{=}{\mathit{a}}^{\mathbf{2}}\left(mod4n\right) \end{gathered}$$

Short answer

$$\begin{gathered} \left(a\right)Itisprovedthat{(\mathrm{n}-\mathrm{a})}^2=a^2(mod\mathrm{n}) \\ \left(b\right){Itisprovedthat(2\mathrm{n}-\mathrm{a})}^2=a^2(mod4\mathrm{n}) \end{gathered}$$

Step-by-step solution

Definition of Congruence

Consider thata, b,n as an integer with $n>0$. If n divides $a-b$, then a is congruent to b modulo n (expressed as$a=b\left(modn\right)$ ).

 Show that (n-a)2=a2(mod n)

a)

Let the term be ${\left(n-a\right)}^2-a^2$. Then,

role="math" localid="1648062351409" $$\begin{gathered} {\left(n-a\right)}^2-a^2=n^2+a^2-2an-a^2 \\ =n^2-2an \\ =n\left(n-2a\right) \end{gathered}$$

It follows that ${\left(n-a\right)}^2-a^2$is divisible by $n$. As a result,${(\mathrm{n}-\mathrm{a})}^2=a^2(\mathrm{mod}\mathrm{n})$.

Thus, it is proved that .role="math" localid="1648062484516" ${(\mathrm{n}-\mathrm{a})}^2=a^2(mod\mathrm{n})$

Show that (2n-a)2=a2(mod 4n)

Let the term be ${\left(2n-a\right)}^2$. Then,

$$\begin{gathered} {\left(2n-a\right)}^2-a^2=4n^2+a^2-4an-a^2 \\ =4n^2-4an \\ =4n\left(n-a\right) \end{gathered}$$

It follows that ${\left(2n-a\right)}^2-a^2$is divisible by $4n$. As a result, ${(2\mathrm{n}-\mathrm{a})}^2=a^2(\mathrm{mod}4\mathrm{n})$.

Thus, it is proved that${(2\mathrm{n}-\mathrm{a})}^2=a^2(\mathrm{mod}4\mathrm{n})$