Show that the square of any positive integer cannot be of the form $(5q + 2)$ or $(5q + 3)$ for any integer $q$.

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Step 1: Let $a$ be any positive integer. Take $b = 5$ as the divisor.
$	herefore a = 5m + r$, $r = 0, 1, 2, 3, 4$
Step 2: Case 1: $a = 5m \Rightarrow a^2 = 25m^2 = 5(5m^2) = 5q$
Case 2: $a = 5m+1 \Rightarrow a^2 = 5(5m^2+2m) + 1 = 5q + 1$
Case 3: $a = 5m+2 \Rightarrow a^2 = 5(5m^2+4m) + 4 = 5q + 4$
Case 4: $a = 5m+3 \Rightarrow a^2 = 5(5m^2+6m+1) + 4 = 5q + 4$
Case 5: $a = 5m+4 \Rightarrow a^2 = 5(5m^2+8m+3) + 1 = 5q + 1$
Step 3: Hence square of any positive integer cannot be of the form $(5q + 2)$ or $(5q + 3)$ for any integer $q$.

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