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Euler's criterion computes $(3 / 726377359)$ by repeated squaring

Example

Let $p = 726377359$ . Repeated squaring computes

$3^{(p - 1) / 2} = 3^{363188679} \equiv - 1 (m o d p),$

and therefore $(3 / p) = - 1$ .

For a checkable transcript, let $q_{i} \equiv 3^{2^{i}} (m o d p)$ and let $c_{i}$ be the accumulated product through bit $i$ of $363188679$ . The rows with bit $0$ leave the accumulator unchanged.

$i$	bit	$q_{i}$	$c_{i}$
0	1	3	3
1	1	9	27
2	1	81	2187
3	0	6561	2187
4	0	43046721	2187
5	0	311816404	2187
6	1	332055490	554374989
7	1	627697086	692682703
8	1	290411863	122079421
9	0	428546868	122079421
10	0	664373569	122079421
11	0	128724442	122079421
12	1	392778933	332499182
13	0	25099053	332499182
14	1	74866315	85056380
15	1	620616191	683996757
16	1	721181380	157844918
17	0	204089129	157844918
18	1	418766369	355227031
19	0	709737354	355227031
20	0	528168097	355227031
21	1	329748334	275223588
22	0	397327928	275223588
23	1	16921694	272602723
24	1	688270323	635018952
25	0	178932138	635018952
26	1	99664525	255823173
27	0	375528119	255823173
28	1	179529638	726377358

Facts & Assumptions

Given: The prime $p = 726377359$ , the binary exponent $363188679$ , and the displayed repeated-squaring transcript.

[L1]

Euler's criterion gives $(a / p) \equiv a^{(p - 1) / 2} (m o d p)$ for every integer $a$ and odd prime $p$ (Euler's criterion: $(a / p) \equiv a^{(p - 1) / 2} (m o d p)$ ).

Verification

technique · direct

1.1

Starting from $q_{0} = 3$ , each table entry satisfies $q_{i + 1} \equiv q_{i}^{2} (m o d p)$ ; the bit column is the binary expansion $363188679 = (10101101001011101000111000111)_{2}$ , and multiplying precisely the $q_{i}$ with bit $1$ gives the displayed accumulators, ending at $726377358$ .

givenalgebra

2.1

The final residue is $726377358 = p - 1 \equiv - 1 (m o d p)$ , and it is not congruent to $1$ because $p > 2$ .

step 1.1algebra

3.1

Applying [L1] with $a = 3$ and using step 2.1 gives $(3 / p) = - 1$ .

L1step 2.1∎

Depends on

Euler's criterion: $(a/p)\equiv a^{(p-1)/2}\pmod p$

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 57 results over 13 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

W. Stein, Elementary Number Theory, Example 4.2.5 (standard reference, not scraped)

Euler's criterion computes (3/726377359)(3/726377359) by repeated squaring