For many fixed $n$, there are nice criteria to check if a prime $p$ can be expressed as $x^2+ny^2$. For example \begin{align} p=x^2+y^2 &\iff p=2\ \vee\ p\equiv 1\pmod 4 \\ p=x^2+2y^2 &\iff p=2\ \vee\ p\equiv 1,3\pmod 8 \\ p=x^2+3y^2 &\iff p=3\ \vee\ p\equiv 1\pmod 3 \\ p=x^2+4y^2 &\iff p\equiv 1\pmod 4 \\ p=x^2+5y^2 &\iff p=5\ \vee\ p\equiv 1,9\pmod {20} \\ p=x^2+6y^2 &\iff p\equiv 1,7\pmod {24} \\ \end{align} What is the generalization for $p$ not prime? The case $n=1$ is well known: A number $a$ is the sum of two squares iff each prime factor of the form $4k+3$ has an even exponent. What is the the general statement for other $n$?
non-primes of the form $x^2+ny^2$
4
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number-theory
elementary-number-theory
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1see here i will hope this will help you! http://maths.dur.ac.uk/gandalf/1213/charlton_primes/charlton_primes_slides.pdf – 2017-02-18
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0Re: Case $n=1$. If $a$ is power of $2$ then $a$ is the sum of two squares iff $a>1$ and $a$ is not a power of $ 4.$ – 2017-02-19
1 Answers
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I wrote about primes for $n \leq 29$ at Primes of the form $x^2+ny^2$
As long as $n$ is idoneal things are not bad. When there are just two genera and favorable conditions, we have the principal form $f = x^2 + n y^2$ and another form $g.$ By Gauss composition, if we have a number represented by either $f$ or $g,$ multiplying by a prime $r = g(x,y)$ takes us to a number represented by the other form. Put another way, multiplying by $r$ takes us back and forth between residues and nonresidues mod some prime that divides $n;$ unless $r|n$ which needs a little more thought.
For $x^2 + 5 y^2,$ the exponent for any prime factor $q \equiv 11,13,17,19 \pmod {20} $ must be even. For primes $q=2$ and $q \equiv 3,7 \pmod {20}, $ the sum of the exponents must be even. These latter primes come from $q = 2 u^2 + 2 uv + 3 v^2.$
jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./bingo
Input three coefficients a b c for positive f(x,y)= a x^2 + b x y + c y^2
1 0 5
Discriminant 20
Maximum number represented?
300
0 =
1 =
4 = 2^2
5 = 5
6 = 2 * 3
9 = 3^2
14 = 2 * 7
16 = 2^4
20 = 2^2 * 5
21 = 3 * 7
24 = 2^3 * 3
25 = 5^2
29 = 29
30 = 2 * 3 * 5
36 = 2^2 * 3^2
41 = 41
45 = 3^2 * 5
46 = 2 * 23
49 = 7^2
54 = 2 * 3^3
56 = 2^3 * 7
61 = 61
64 = 2^6
69 = 3 * 23
70 = 2 * 5 * 7
80 = 2^4 * 5
81 = 3^4
84 = 2^2 * 3 * 7
86 = 2 * 43
89 = 89
94 = 2 * 47
96 = 2^5 * 3
100 = 2^2 * 5^2
101 = 101
105 = 3 * 5 * 7
109 = 109
116 = 2^2 * 29
120 = 2^3 * 3 * 5
121 = 11^2
125 = 5^3
126 = 2 * 3^2 * 7
129 = 3 * 43
134 = 2 * 67
141 = 3 * 47
144 = 2^4 * 3^2
145 = 5 * 29
149 = 149
150 = 2 * 3 * 5^2
161 = 7 * 23
164 = 2^2 * 41
166 = 2 * 83
169 = 13^2
174 = 2 * 3 * 29
180 = 2^2 * 3^2 * 5
181 = 181
184 = 2^3 * 23
189 = 3^3 * 7
196 = 2^2 * 7^2
201 = 3 * 67
205 = 5 * 41
206 = 2 * 103
214 = 2 * 107
216 = 2^3 * 3^3
224 = 2^5 * 7
225 = 3^2 * 5^2
229 = 229
230 = 2 * 5 * 23
241 = 241
244 = 2^2 * 61
245 = 5 * 7^2
246 = 2 * 3 * 41
249 = 3 * 83
254 = 2 * 127
256 = 2^8
261 = 3^2 * 29
269 = 269
270 = 2 * 3^3 * 5
276 = 2^2 * 3 * 23
280 = 2^3 * 5 * 7
281 = 281
289 = 17^2
294 = 2 * 3 * 7^2
1 0 5
jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$
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For $x^2 + 6 y^2,$ the exponent for any prime factor $q \equiv 13,17,19,23 \pmod {24} $ must be even. For primes $q=2,3$ and $q \equiv 5,11 \pmod {24}, $ the sum of the exponents must be even. These latter primes come from $q = 2 u^2 + 3 v^2.$
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jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./bingo
Input three coefficients a b c for positive f(x,y)= a x^2 + b x y + c y^2
1 0 6
Discriminant 24
Maximum number represented?
300
0 =
1 =
4 = 2^2
6 = 2 * 3
7 = 7
9 = 3^2
10 = 2 * 5
15 = 3 * 5
16 = 2^4
22 = 2 * 11
24 = 2^3 * 3
25 = 5^2
28 = 2^2 * 7
31 = 31
33 = 3 * 11
36 = 2^2 * 3^2
40 = 2^3 * 5
42 = 2 * 3 * 7
49 = 7^2
54 = 2 * 3^3
55 = 5 * 11
58 = 2 * 29
60 = 2^2 * 3 * 5
63 = 3^2 * 7
64 = 2^6
70 = 2 * 5 * 7
73 = 73
79 = 79
81 = 3^4
87 = 3 * 29
88 = 2^3 * 11
90 = 2 * 3^2 * 5
96 = 2^5 * 3
97 = 97
100 = 2^2 * 5^2
103 = 103
105 = 3 * 5 * 7
106 = 2 * 53
112 = 2^4 * 7
118 = 2 * 59
121 = 11^2
124 = 2^2 * 31
127 = 127
132 = 2^2 * 3 * 11
135 = 3^3 * 5
144 = 2^4 * 3^2
145 = 5 * 29
150 = 2 * 3 * 5^2
151 = 151
154 = 2 * 7 * 11
159 = 3 * 53
160 = 2^5 * 5
166 = 2 * 83
168 = 2^3 * 3 * 7
169 = 13^2
175 = 5^2 * 7
177 = 3 * 59
186 = 2 * 3 * 31
193 = 193
196 = 2^2 * 7^2
198 = 2 * 3^2 * 11
199 = 199
202 = 2 * 101
214 = 2 * 107
216 = 2^3 * 3^3
217 = 7 * 31
220 = 2^2 * 5 * 11
223 = 223
225 = 3^2 * 5^2
231 = 3 * 7 * 11
232 = 2^3 * 29
240 = 2^4 * 3 * 5
241 = 241
249 = 3 * 83
250 = 2 * 5^3
252 = 2^2 * 3^2 * 7
256 = 2^8
262 = 2 * 131
265 = 5 * 53
271 = 271
279 = 3^2 * 31
280 = 2^3 * 5 * 7
289 = 17^2
292 = 2^2 * 73
294 = 2 * 3 * 7^2
295 = 5 * 59
297 = 3^3 * 11
298 = 2 * 149
1 0 6
jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$
==============================
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For $x^2 + 10 y^2,$ the exponent for any prime factor $q \equiv 3, 17, 21, 27, 29, 31, 33, 39 \pmod {40} $ must be even. For primes $q=2,5$ and $q \equiv 7, 13, 23, 37 \pmod {40}, $ the sum of the exponents must be even. These latter primes come from $q = 2 u^2 + 5 v^2.$
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jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./bingo
Input three coefficients a b c for positive f(x,y)= a x^2 + b x y + c y^2
1 0 10
Discriminant 40
Maximum number represented?
300
0 =
1 =
4 = 2^2
9 = 3^2
10 = 2 * 5
11 = 11
14 = 2 * 7
16 = 2^4
19 = 19
25 = 5^2
26 = 2 * 13
35 = 5 * 7
36 = 2^2 * 3^2
40 = 2^3 * 5
41 = 41
44 = 2^2 * 11
46 = 2 * 23
49 = 7^2
56 = 2^3 * 7
59 = 59
64 = 2^6
65 = 5 * 13
74 = 2 * 37
76 = 2^2 * 19
81 = 3^4
89 = 89
90 = 2 * 3^2 * 5
91 = 7 * 13
94 = 2 * 47
99 = 3^2 * 11
100 = 2^2 * 5^2
104 = 2^3 * 13
106 = 2 * 53
110 = 2 * 5 * 11
115 = 5 * 23
121 = 11^2
126 = 2 * 3^2 * 7
131 = 131
139 = 139
140 = 2^2 * 5 * 7
144 = 2^4 * 3^2
154 = 2 * 7 * 11
160 = 2^5 * 5
161 = 7 * 23
164 = 2^2 * 41
169 = 13^2
171 = 3^2 * 19
176 = 2^4 * 11
179 = 179
184 = 2^3 * 23
185 = 5 * 37
190 = 2 * 5 * 19
196 = 2^2 * 7^2
206 = 2 * 103
209 = 11 * 19
211 = 211
224 = 2^5 * 7
225 = 3^2 * 5^2
234 = 2 * 3^2 * 13
235 = 5 * 47
236 = 2^2 * 59
241 = 241
250 = 2 * 5^3
251 = 251
254 = 2 * 127
256 = 2^8
259 = 7 * 37
260 = 2^2 * 5 * 13
265 = 5 * 53
266 = 2 * 7 * 19
275 = 5^2 * 11
281 = 281
286 = 2 * 11 * 13
289 = 17^2
296 = 2^3 * 37
299 = 13 * 23
1 0 10
jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$
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0Thanks a lot. The idoneal case is absolutely sufficient for me right now. Though I don't really see how you get to your result – 2017-02-18
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0ah, okay that makes it clearer. I think that will be enough for me to figure out the details. – 2017-02-18