Question 1141725
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The problem as you pose it is trivial.  "This sequence" is the three numbers 2, 5, and 8.  None of 2, 5, or 8 is a perfect square.<br>
The problem you intended to pose is an interesting one, involving the infinite arithmetic sequence 2, 5, 8, ....<br>
The numbers in this sequence are the numbers of the form 3n+2, where n is an integer.  We need to show that there are no squares of integers that are of that form.<br>
Every integer can be represented in exactly one way as either 3k-1, 3k, or 3k+1, where k is an integer.<br>
If the integer is of the form 3k-1 or 3k+1, then the square of the integer is of the form 3n+1:
{{{(3k-1)^2 = 9k^2-6k+1 = 3(3k^2-2k)+1}}}
{{{(3k+1)^2 = 9k^2+6k+1 = 3(3k^2+2k)+1}}}<br>
And if the integer is of the form 3k, then the square of the integer is of the form 3n:
{{{(3k)^2 = 9k^2 = 3(3k^2)}}}<br>
So the square of any integer is either of the form 3n or 3n+1 -- never 3n+2.