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prove that if in an arithmetic sequence of natural numbers, one of the terms is a perfect square, then there are many terms which are perfect squares. Is there an arithmetic sequence of natural numbers in which no terms term is a perfect square ?
The nth term of an arithmetic sequence is tn = t1 + (n-1)d
second term ,t2 = t1 + (2-1)d
implies t2 = t1 + d
ie, d = t2 - t1
d is an integer ,d cannot be negative
If  d=0

Then the terms are all the same ,and if any one of them is a perfect square ,then all the numbers are perfect squares.
If d is a natural numbers

Suppose the kth term is a perfect square ,p2
ie, tk= t1 + (k-1)d = p2
add 2pmd + m2d2 to both sides where m is a natural number.
t1 + (k-1)d +2pmd + m2d2     =      p2   +2pmd +m2d2 
Take d as a commn factor from the last three terms on the left and factor the right side as a perfect square
t1 +[ (k-1) +2pm + m2d ]d    =      [p+md]2
Here right side is a perfect square for infinitely many values of m.
Hence the theorem is proved.
Qn:Is there an arithmetic sequence of natural numbers in which no terms term is a perfect square ?
Ans:
A simple case with d = 0,
2,2,2,......
This one doesn't contain any term that is a perfect square
Consider the sequence 2,6,10,14,18,22,....
 This sequence contains only even terms.
If this sequence contained a term which is a perfect square it would be even.
But every even perfect square is divisible by 4.
This sequence has t1 = 2 ,d = 4
Thus its mth term is tn = t1 +(n-1)d = 2 + (n-1)4
                                                        = 4n - 2
 If we divide4n -2 by 4 we get n - 1/2 which is not an integer.So the sequence cannot contain a perfect square


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