# prove that if an arithmetic progression of natural numbers,one of the term is a perfect square,then there are many terms which are perfect squares.Is there an arithmetic progression of natural numbers in which no term is a perfect square?

a) Prove that if an arithmetic progression of natural numbers,one of the term is a perfect square,then there are many terms which are perfect squares.

Proof: As we know that

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

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

Case 1:  Let 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.

b)Is there an arithmetic progression of natural numbers in which no term is a perfect square

Proof:
Consider the sequence:  2,6,10,14,18,22,...

Because this sequence contains only even terms.

If this sequence contained a term that was a perfect square, it would be even.

As 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-(1/2) which is not  an integer.

So,  the sequence cannot contain a perfect square.

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