<s>[INST] 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.
 [/INST] 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 $$3^{2^{n+1}} – 1=X^2+Y^2$$
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