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Given 2, 5, 8 as a sequence. Prove that none of the terms 
is a perfect square
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Please notice how I edited your post to create an adequate meaning.
This sequence is an arithmetic progression with the common difference of 3 and the first term 2.
When dividing by 3, all the terms of the progression give the remainder of 2.
No one perfect square gives the remainder 2 when is dividing by 3.
Indeed, let's consider an arbitrary square 
of an integer number "n".
If n is multiple of 3, n = 3k, then is a multiple of 9; hence, it is a multiple of 3, giving the remainder 0 whin divided by 3.
If n = 3k+1, then = 
, giving the remainder 1, when divided by 3.
If n = 3k+2, then = 
, giving the remainder 1, when divided by 3.
It proves the statement.
