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I am looking for all two consecutive integers A and A+1,which can be represented as sums of two squares $A=a^2+b^2$ and $A+1=c^2+d^2$, $a,b,c,d>0$

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There are infinitely many such $A$, and it is unlikely that there is any useful characterization. –  Gerry Myerson Jul 8 '13 at 12:25
    
Can be given a formula for a certain group of such numbers –  Rally Jul 8 '13 at 12:28
    
A related question was discussed at math.stackexchange.com/questions/46451/… but the squares were allowed to be zero, which makes life easier. –  Gerry Myerson Jul 8 '13 at 12:33
    
A simple infinite family would be $A = (n^2+1)^2$. –  Daniel Fischer Jul 8 '13 at 12:34

1 Answer 1

Not a complete answer --- I doubt there is one --- but $$(n^2-n)^2+(n^2-n)^2,(n^2-2n)^2+(n^2-1)^2,(n^2-n-1)^2+(n^2-n+1)^2$$ gives three consecutive numbers, each a sum of two non-zero squares. This example is taken from Cochrane and Dressler, Consecutive triples of sums of two squares, which also cites earlier results about consecutive pairs of sums of two squares.

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