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 08 '13 at 12:25
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Can be given a formula for a certain group of such numbers – Rally Jul 08 '13 at 12:28
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A related question was discussed at http://math.stackexchange.com/questions/46451/three-consecutive-sums-of-two-squares but the squares were allowed to be zero, which makes life easier. – Gerry Myerson Jul 08 '13 at 12:33
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A simple infinite family would be $A = (n^2+1)^2$. – Daniel Fischer Jul 08 '13 at 12:34
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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.
Gerry Myerson
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Take a member of the modular group, i.e. a matrix with integer entries and determinant one. Let's call it M={{a,b},{c,d}}. There are plenty of such matrices, in fact factorization of any two consecutive integers is a good source of these beasts (it is a kind of poetic justice that any two consecutive integers will give you your special consecutive integers - or half integers).
For any such matrix consider these two integers (or half integers)
$X = (\frac{a+d}{2})^2+(\frac{c-b}{2})^2$
$Y = (\frac{a-d}{2})^2+(\frac{c+b}{2})^2$
Elements of the modular group come in two flavors, they have either one even number among its entries or two. The latter case will give you the desired integers. The fact that they are consecutive follows from the det(M)=1 condition (the former case will give you some interesting solutions with half integers, which you may donate at the Salvation Army).
Tamref Relue
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