Question

If $a\in \mathrm{\mathbb{Z}}$, prove that $a^2$ is not congruent to 2 modulo 4 or to 3 modulo 4.

Short answer

It is proved that $a^2$ is not congruent to 2 modulo 4 or to 3 modulo 4.

Step-by-step solution

Definition of Congruence 

Consider a, b, n as an integer with $n>0$. If n divides $a-b$ then a iscongruent tob modulo n (expressed as $a\equiv b\left(modn\right)$).

Show that a2 is not congruent to 2 modulo 4 or to 3 modulo 4

Every number $a\in \mathrm{\mathbb{Z}}$ would be of the form $4k,4k+1,4k+2,4k+3$.

When $a=4k$ then it is as follows:

${\left(4k\right)}^2\equiv 0\left(mod4\right)$

When $a=4k+1$ then it is as follows:

role="math" localid="1646224843178" $\begin{array}{rcl}{\left(4k+1\right)}^2& =& 4\left(4k^2+2k\right)+1\\ & \equiv & 1\left(mod4\right)\end{array}$

When $a=4k+2$ then it is as follows:

role="math" localid="1646224860994" $\begin{array}{rcl}{\left(4k+2\right)}^2& =& 4{\left(2k+1\right)}^2\\ & \equiv & 0\left(mod4\right)\end{array}$

When $a=4k+3$ then it is as follows:

$\begin{array}{rcl}{\left(4k+3\right)}^2& =& 4\left(4k^2+6k+2\right)+1\\ & \equiv & 1\left(mod4\right)\end{array}$

2 modulo 4 and 3 modulo 4 are notcongruent toany of the squares.

Thus, it is proved that$a^2$ is notcongruent to2 modulo 4 or to 3 modulo 4.