$\text{a chia 4 dư 1}$
$\text{⇒a=4k+1}$
$\text{b chia 4 dư 3}$
$\text{⇒b=4n+3}$
$\text{b²-a²}$
$\text{=(4n+3)²-(4k+1)²}$
$\text{=16n²+24n+9-(16k²+8k+1)}$
$\text{=16n²+24n+9-16k²-8k-1}$
$\text{=24n-8k+8}$
$\text{=4(6n-2k+2)}$
$\text{4 chia hết cho 4}$
$\text{⇒4(6n-2k+2) chia hết cho 4}$
$\text{Hay b²-a² chia hết cho 4}$