Question 3
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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Solution

Let a be an arbitrary positive integer.
Then, by Euclid's division algorithm, corresponding to the positive integers a and 5, there exist non - negative integers m and r such that:
$\Rightarrow a = 5 m + r, w h e r e 0 \leq r < 5$
$\Rightarrow a^{2} = (5 m + r)^{2} = 25 m^{2} + r^{2} + 10 m r [∵ (a + b)^{2} = a^{2} + 2 a b + b^{2}]$
$\Rightarrow a^{2} = 5 (5 m^{2} + 2 m r) + r^{2}$ ...(i)
Where, $0 \leq r < 5$

Case 1
When r = 0, putting r = 0 in eq. (i) we get,
$a^{2} = 5 (5 m^{2}) = 5 q$
where, $q = 5 m^{2}$ is an integer.

case II
When r = 1, putting r = 1 is eq. (i) we get,
$a^{2} = 5 (5 m^{2} + 2 m) + 1$
$\Rightarrow a^{2} = 5 q + 1$
Where, $q = (5 m^{2} + 2 m)$ is an integer.

Case III
When r = 2, putting r = in eq. (i) we get,
$a^{2} = 5 (5 m^{2} + 4 m) + 4 = 5 q + 4$
Where, $q = (5 m^{2} + 4 m) + 4 = 5 q + 4$

Case IV
When r = 3, putting r = 3 in eq (i) we get,
$a^{2} = 5 (5 m^{2} + 6 m) + 9 = 5 (5 m^{2} + 6 m) + 5 + 4$
$= 5 (5 m^{2} + 6 m + 1) + 4 = 5 q + 4$
Where, $q = (5 m^{2} + 6 m + 1)$ is an integer.

Case V
When r = 4, putting r = 4 in eq (i), we get,
$a^{2} = 5 (5 m^{2} + 8 m) + 16 = 5 (5 m^{2} + 8 m) + 15 + 1$
$\Rightarrow a^{2} = 5 (5 m^{2} + 8 m + 3) + 1 = 5 q + 1$
Where, $q = (5 m^{2} + 8 m + 3)$ is an integer.
Hence, the square of any positive integer cannot be of the form 5q + 2 or 5q + 3 for any integer q.

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