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? 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.

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 $$3^{2^{n+1}} – 1=X^2+Y^2$$
