Let $n$ be a square-free positive integer and denote by $S_n(a, b, c)$ the number of solutions in integers, $x, y,z$, of the equation
$ax^2 + by^2 + cz^2 = n$.
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Then a necessary condition for $n$ to be a congruent number is that
$S_n(2, 1, 8) = 2S_n(2, 1, 32)$ $n$ odd
and
$S_n(8, 2, 16) = 2S_n(8, 2, 64)$ $n$ even.
Moreover, if the Birch and Swinnerton-Dyer conjecture is true then this condition is also sufficient

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