SOLUTION: Prove that in an arithmetis sequence of natural numbers, one of the terms is aperfect square , then there are many terms which are perfect squares. Is there an arithmetic sequence

Proof:

The nth term of an arithmetic sequence is

an = a1 + (n-1)d

Suppose this is a sequence of all natural numbers.

Then a1 and a2 are natural numbers

a2 = a1 + (2-1)d

a2-a1 = d

Therefore d is an integer.  d cannot be negative, otherwise
there would eventally be negative terms in the sequence.

Case 1:
d = 0.  Then the terms are all the same, and if any one
of them is a perfect square, then they all are, and the 
theorem is proved.

Case 2:
d is a natural number.

Suppose the kth term is a perfect square p²

ak = 

           a1 + (k-1)d = p²

add 2pmd + m²d² to both sides where m is a natural number

a1 + (k-1)d + 2pmd + m²d² = p² + 2pmd + m²d²

Factor d out of the last three terms on the left
and factor the right side as a perfect square:

     a1 + [(k-1)+2pm+m²]d = (p+md)²

The right side is a perfect square, and the left side
is the (k-1+2pm+m²)th term for infinitely many values of m.
 
Thus the theorem is proved.

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A simple case is with d=0

2,2,2,2,2,2,2,...

But your teacher may not go for that as it is too trivial.

However this one doesn't contain any term that is a perfect square: 

2,6,10,14,18,22,...

Here's why.  This sequence contains only even terms. If this 
sequence contained a term that was a perfect square, it would 
be even.  But every even perfect square is divisible by 4.

This sequence has a1=2, d=4, thus its nth term is: 

an = a1+(n-1)d = 2+(n-1)*4 = 2+4n-4 = 4n-2.

But if we divide 4n-2 by 4 we get n- which is not 
an integer.  So the sequence cannot contain a perfect square.

Edwin