Re: Proving RC6' quadratic permutation is a permutation

From the success-when-admitting-defeat department:

* Marius Bernklev

Out of idle curiosity, I wonder how exactly I would prove, or even
"just" convince myself, that RC6' quadratic permutation, A×(2×A + 1)
(mod 2^N) actually is a permutation.

The only thing I see so far is that for
A×(2×A + 1) = B×(2×B + 1) (mod 2^N),

A and B would have to be of equal parity.

Why do A and B need to be equal?

A×(2×A + 1) = B×(2×B + 1)

2×A² + A = 2×B² + B

2×A² - 2xB² + A - B = 0

2×(A² - B²) + A - B = 0

2×(A + B)×(A - B) + (A - B) = 0

(A - B) × (2×(A + B) + 1) = 0
^^^^^^^^^^^^^^^
is obviously not 0, so

A - B = 0

A = B

Does this seem right?


--
Marius Bernklev

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