Question

In an arithmetic sequence having terms natural numbers, prove that if one of
the terms is a perfect square, it will have more that this as the perfect square
term​

Answer 1

Step-by-step explanation:

Prove that if an arithmetis sequence of natural numbers, one of the terms is a perfect square, then there are many terms which are perfect squares.

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.