show that the square of any positive integer cannot be of the form 6 m + 2 or 6 m + 5 for any integer m

Let  a  be any positive integer.

By Euclid's division lemma ,

 a = bm + r  ,  where  b = 6

⇒  a = 6m + r

 where  r  can be 0, 1, 2, 3, 4, 5.

∴  a = 6m  if  r = 0

 a = 6m + 1  if  r = 1

 a = 6m + 2  if  r = 2

 a = 6m + 3  if  r = 3

 a = 6m + 4  if  r = 4

 a = 6m + 5  if  r = 5

$$\begin{gathered} \mathrm{CASE} 1 : \\ \mathrm{When} \mathrm{a} = 6\mathrm{m} \\ \Rightarrow {\mathrm{a}}^2 = {\left(6\mathrm{m}\right)}^2 = 36{\mathrm{m}}^2 = 6\left(6{\mathrm{m}}^2\right) = 6\mathrm{p} \\ \mathrm{CASE} 2 : \\ \mathrm{When} \mathrm{a} = 6\mathrm{m}+1 \\ \Rightarrow {\mathrm{a}}^2 = {\left(6\mathrm{m}+1\right)}^2 = 36{\mathrm{m}}^2+1+12\mathrm{m} = 6\left(6{\mathrm{m}}^2+2\mathrm{m}\right) + 1 = 6\mathrm{p} + 1 \\ \mathrm{CASE} 3 : \\ \mathrm{When} \mathrm{a} = 6\mathrm{m} + 2 \\ \Rightarrow {\mathrm{a}}^2 = {\left(6\mathrm{m}+2\right)}^2 = 36{\mathrm{m}}^2+4+24\mathrm{m} = 6\left(6{\mathrm{m}}^2+4\mathrm{m}\right) + 4 = 6\mathrm{p} + 4 \\ \mathrm{CASE} 4 : \\ \mathrm{When} \mathrm{a} = 6\mathrm{m} + 3 \\ \Rightarrow {\mathrm{a}}^2 = {\left(6\mathrm{m}+3\right)}^2 = 36{\mathrm{m}}^2+9+36\mathrm{m} = 6\left(6{\mathrm{m}}^2+6\mathrm{m} + 1\right) + 3 = 6\mathrm{p} + 3 \\ \mathrm{CASE} 5 : \\ \mathrm{When} \mathrm{a} = 6\mathrm{m}+4 \\ \Rightarrow {\mathrm{a}}^2 = {\left(6\mathrm{m}+4\right)}^2 = 36{\mathrm{m}}^2+16+48\mathrm{m} = 6\left(6{\mathrm{m}}^2+8\mathrm{m} + 2\right) + 4 = 6\mathrm{p} + 4 \\ \mathrm{CASE} 6 : \\ \mathrm{When} \mathrm{a} = 6\mathrm{m} + 5 \\ \Rightarrow {\mathrm{a}}^2 = {\left(6\mathrm{m}+5\right)}^2 = 36{\mathrm{m}}^2+25+30\mathrm{m} = 6\left(6{\mathrm{m}}^2+5\mathrm{m} + 4\right) + 1 = 6\mathrm{p} + 1 \end{gathered}$$
From all these cases, it is clear that square of any positive integer can not be of the form of 
6m + 2 or 6m + 5.

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