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Quadratic reciprocity
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===Factoring ''n''<sup>2</sup> β 5=== Consider the polynomial <math>f(n) = n^2 - 5</math> and its values for <math>n \in \N.</math> The prime factorizations of these values are given as follows: {| class="wikitable" style="text-align:right;" |- ! ''n'' !! colspan=2 | {{tmath|f(n)}} || !! ''n'' !! colspan=2 | {{tmath|f(n)}} || !! ''n'' !! colspan=2 | {{tmath|f(n)}} |- | 1 || β4 || β2<sup>2</sup> || || 16 || 251 || 251 || || 31 || 956 || 2<sup>2</sup>⋅239 |- | 2 || β1 || β1 || || 17 || 284 || 2<sup>2</sup>⋅71 || || 32 || 1019 || 1019 |- | 3 || 4 || 2<sup>2</sup> || || 18 || 319 || 11⋅29 || || 33 || 1084 || 2<sup>2</sup>⋅271 |- | 4 || 11 || 11 || || 19 || 356 || 2<sup>2</sup>⋅89 || || 34 || 1151 || 1151 |- | 5 || 20 || 2<sup>2</sup>⋅5 || || 20 || 395 || 5⋅79 || || 35 || 1220 || 2<sup>2</sup>⋅5⋅61 |- | 6 || 31 || 31 || || 21 || 436 || 2<sup>2</sup>⋅109 || || 36 || 1291 || 1291 |- | 7 || 44 || 2<sup>2</sup>⋅11 || || 22 || 479 || 479 || || 37 || 1364 || 2<sup>2</sup>⋅11⋅31 |- | 8 || 59 || 59 || || 23 || 524 || 2<sup>2</sup>⋅131 || || 38 || 1439 || 1439 |- | 9 || 76 || 2<sup>2</sup>⋅19 || || 24 || 571 || 571 || || 39 || 1516 || 2<sup>2</sup>⋅379 |- | 10 || 95 || 5⋅19 || || 25 || 620 || 2<sup>2</sup>⋅5⋅31 || || 40 || 1595 || 5⋅11⋅29 |- | 11 || 116 || 2<sup>2</sup>⋅29 || || 26 || 671 || 11⋅61 || || 41 || 1676 || 2<sup>2</sup>⋅419 |- | 12 || 139 || 139 || || 27 || 724 || 2<sup>2</sup>⋅181 || || 42 || 1759 || 1759 |- | 13 || 164 || 2<sup>2</sup>⋅41 || || 28 || 779 || 19⋅41 || || 43 || 1844 || 2<sup>2</sup>⋅461 |- | 14 || 191 || 191 || || 29 || 836 || 2<sup>2</sup>⋅11⋅19 || || 44 || 1931 || 1931 |- | 15 || 220 || 2<sup>2</sup>⋅5⋅11 || || 30 || 895 || 5⋅179 || || 45 || 2020 || 2<sup>2</sup>⋅5⋅101 |} The prime factors <math>p</math> dividing <math>f(n)</math> are <math>p=2,5</math>, and every prime whose final digit is <math>1</math> or <math>9</math>; no primes ending in <math>3</math> or <math>7</math> ever appear. Now, <math>p</math> is a prime factor of some <math>n^2-5</math> whenever <math>n^2 - 5 \equiv 0 \bmod p</math>, i.e. whenever <math>n^2 \equiv 5 \bmod p,</math> i.e. whenever 5 is a quadratic residue modulo <math>p</math>. This happens for <math>p= 2, 5</math> and those primes with <math>p\equiv 1, 4 \bmod 5,</math> and the latter numbers <math>1=(\pm1)^2</math> and <math>4=(\pm2)^2</math> are precisely the quadratic residues modulo <math>5</math>. Therefore, except for <math>p = 2,5</math>, we have that <math>5</math> is a quadratic residue modulo <math>p</math> iff <math>p</math> is a quadratic residue modulo <math>5</math>. The law of quadratic reciprocity gives a similar characterization of prime divisors of <math>f(n)=n^2 - q</math> for any prime ''q'', which leads to a characterization for any integer <math>q</math>.
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