How to show that that the following three consecutive numbers $3^{2^{10}} − 1$, $3^{2^{10}}$,$3^{2^{10}}+ 1$ are the sum of two squares?

Question

Show that the following three consecutive numbers: $$ 3^{2^{10}} − 1, 3^{2^{10}} , 3^{2^{10}} + 1 $$ can be represented as sums of two integer squares.

Answer 1

It is known that the set of sums of two squares is closed under the multiplication. On the other hand, it is clear that $3^{2^{10}}$ and $3^{2^{10}} + 1$ satisfy the statement.

Let us prove that in general $3^{2^{n}} – 1$ is a sum of two squares.

By induction, we have $3^{2^{0}} – 1=2=1+1$ so assume that $$3^{2^{n}} – 1=x^2+y^2$$ We have $$3^{2^{n+1}} – 1=(3^{2^{n}} + 1)( 3^{2^{n}} – 1)$$ It follows $$3^{2^{n+1}} – 1=(3^{2^{n}} + 1)( 3^{2^{n}} – 1)= (3^{2^{n}} + 1)( x^2+y^2)$$ The first factor is obviously a sum of two squares, therefore $$3^{2^{n+1}} – 1=X^2+Y^2$$

Answer 2

the second and third can be expressed as $(3^{2^{9}})^2+0^2$ and $(3^{2^9})^2+1^2$

For the first:

$3^{2^{10}}-1=$ $(3^{2^{9}}+1)(3^{2^8}+1)(3^{2^7}+1)(3^{2^6}+1)(3^{2^5}+1)(3^{2^4}+1)(3^{2^3}+1)(3^{2^2}+1)(3^{2}+1)8$

Each of the factors is trivially a sum of two squares, So all we need to conclude is the fact that the product of two numbers that are the sum of two squares is also the sum of two squares, this is immediate by the Brahmagupta–Fibonacci identity.