Let's consider an arithmetical or geometrical progression with all elements natural numbers, which starts with a perfect square. Prove that the progression includes an infinity of perfect squares!

lets start with the geometric progression

given a starting value of x^2 and a constant ratio of r we want to find t such that x^2*r^t is a perfect square.  Since all terms must be natural numbers then we simply choose t=2k and we have x^2*r^t=x^2*r^(2k)=(x*r^k)^2=perfect square

for the arithmetic progresson

given a starting value of x^2 and a constant difference of d we want to find t such that x^2+t*d is a prefect square, say y.  then

x^2+t*d=y^2

t*d=y^2-x^2

t*d=(y+x)(y-x)

t=(y+x)(y-x)/d

now for any k>0 set y=x+d*k then we have

t=(2x+d*k)*(d*k)/d=k*(2x+d*k)

thus x^2+t*d=x^2+2*k*d*x+d^2*k^2=(x+dk)^2

and thus for any x,d or x,r defineing either a arithmetic or geometric progresson I can find an infinite number of t such hat the t'th member of that progression is a perfect square

Posted by Daniel on 2007-12-20 12:21:09