## Gauss’s Criterion

### May 1, 2018

This isn’t hard, though I admit I mixed up b and p the first time I did it:

```(define (gauss-criterion b p)
(define (gauss 2k-1) (modulo (* 2k-1 b) p))
(expt -1 (length (filter even? (filter positive? (map gauss (range 1 p 2)))))))```

And here is a demonstration that it correctly computes the Legendre symbol:

```> (do ((ps (cdr (primes 100)) (cdr ps))) ((null? ps))
(do ((b 1 (+ b 1))) ((= (car ps) b))
(assert (gauss-criterion b (car ps))
(jacobi b (car ps)))))```

No news is good news. You can run the program at https://ideone.com/ONFr06.

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### One Response to “Gauss’s Criterion”

1. Daniel said

Here’s a solution in Python 2.

```def legendre(b, p):
l = (b ** ((p - 1) / 2)) % p
if l == p - 1:
l = -1
return l

def gauss(b, p):
r = ((x * b) % p for x in xrange(1, p, 2))
t = sum(1 for x in r if (x % 2) == 0)
return (-1) ** t

ps = [3, 5, 7, 11, 13, 17, 19, 23, 29]
for p in ps:
for b in range(1, p):
assert gauss(b, p) == legendre(b, p)
```