5)
If x
and y are integers, then x2 + xy + y2 can never be
negative, and x2 + xy + y2 = 0 only if both x = 0 and y =
0.
If x and y
are integers, then x2 + xy + y2 can never be negative,
and x2 + xy + y2 = 0 only if both x = 0 and y = 0. This is true even when considering negative
numbers.
-10
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-9
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-8
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-7
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-6
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-5
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-4
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-3
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-2
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-1
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0
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1
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2
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3
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4
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5
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6
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7
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8
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9
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10
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||
10
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100
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91
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84
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79
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76
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75
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76
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79
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84
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91
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100
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111
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124
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139
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156
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175
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196
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219
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244
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271
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300
|
10
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9
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91
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81
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73
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67
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63
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61
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61
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63
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67
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73
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81
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91
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103
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117
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133
|
151
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171
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193
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217
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243
|
271
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9
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8
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84
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73
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64
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57
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52
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49
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48
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49
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52
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57
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64
|
73
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84
|
97
|
112
|
129
|
148
|
169
|
192
|
217
|
244
|
8
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7
|
79
|
67
|
57
|
49
|
43
|
39
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37
|
37
|
39
|
43
|
49
|
57
|
67
|
79
|
93
|
109
|
127
|
147
|
169
|
193
|
219
|
7
|
6
|
76
|
63
|
52
|
43
|
36
|
31
|
28
|
27
|
28
|
31
|
36
|
43
|
52
|
63
|
76
|
91
|
108
|
127
|
148
|
171
|
196
|
6
|
5
|
75
|
61
|
49
|
39
|
31
|
25
|
21
|
19
|
19
|
21
|
25
|
31
|
39
|
49
|
61
|
75
|
91
|
109
|
129
|
151
|
175
|
5
|
4
|
76
|
61
|
48
|
37
|
28
|
21
|
16
|
13
|
12
|
13
|
16
|
21
|
28
|
37
|
48
|
61
|
76
|
93
|
112
|
133
|
156
|
4
|
3
|
79
|
63
|
49
|
37
|
27
|
19
|
13
|
9
|
7
|
7
|
9
|
13
|
19
|
27
|
37
|
49
|
63
|
79
|
97
|
117
|
139
|
3
|
2
|
84
|
67
|
52
|
39
|
28
|
19
|
12
|
7
|
4
|
3
|
4
|
7
|
12
|
19
|
28
|
39
|
52
|
67
|
84
|
103
|
124
|
2
|
1
|
91
|
73
|
57
|
43
|
31
|
21
|
13
|
7
|
3
|
1
|
1
|
3
|
7
|
13
|
21
|
31
|
43
|
57
|
73
|
91
|
111
|
1
|
0
|
100
|
81
|
64
|
49
|
36
|
25
|
16
|
9
|
4
|
1
|
0
|
1
|
4
|
9
|
16
|
25
|
36
|
49
|
64
|
81
|
100
|
0
|
-1
|
111
|
91
|
73
|
57
|
43
|
31
|
21
|
13
|
7
|
3
|
1
|
1
|
3
|
7
|
13
|
21
|
31
|
43
|
57
|
73
|
91
|
-1
|
-2
|
124
|
103
|
84
|
67
|
52
|
39
|
28
|
19
|
12
|
7
|
4
|
3
|
4
|
7
|
12
|
19
|
28
|
39
|
52
|
67
|
84
|
-2
|
-3
|
139
|
117
|
97
|
79
|
63
|
49
|
37
|
27
|
19
|
13
|
9
|
7
|
7
|
9
|
13
|
19
|
27
|
37
|
49
|
63
|
79
|
-3
|
-4
|
156
|
133
|
112
|
93
|
76
|
61
|
48
|
37
|
28
|
21
|
16
|
13
|
12
|
13
|
16
|
21
|
28
|
37
|
48
|
61
|
76
|
-4
|
-5
|
175
|
151
|
129
|
109
|
91
|
75
|
61
|
49
|
39
|
31
|
25
|
21
|
19
|
19
|
21
|
25
|
31
|
39
|
49
|
61
|
75
|
-5
|
-6
|
196
|
171
|
148
|
127
|
108
|
91
|
76
|
63
|
52
|
43
|
36
|
31
|
28
|
27
|
28
|
31
|
36
|
43
|
52
|
63
|
76
|
-6
|
-7
|
219
|
193
|
169
|
147
|
127
|
109
|
93
|
79
|
67
|
57
|
49
|
43
|
39
|
37
|
37
|
39
|
43
|
49
|
57
|
67
|
79
|
-7
|
-8
|
244
|
217
|
192
|
169
|
148
|
129
|
112
|
97
|
84
|
73
|
64
|
57
|
52
|
49
|
48
|
49
|
52
|
57
|
64
|
73
|
84
|
-8
|
-9
|
271
|
243
|
217
|
193
|
171
|
151
|
133
|
117
|
103
|
91
|
81
|
73
|
67
|
63
|
61
|
61
|
63
|
67
|
73
|
81
|
91
|
-9
|
-10
|
300
|
271
|
244
|
219
|
196
|
175
|
156
|
139
|
124
|
111
|
100
|
91
|
84
|
79
|
76
|
75
|
76
|
79
|
84
|
91
|
100
|
-10
|
-10
|
-9
|
-8
|
-7
|
-6
|
-5
|
-4
|
-3
|
-2
|
-1
|
0
|
1
|
2
|
3
|
4
|
5
|
6
|
7
|
8
|
9
|
10
|
The fact
that x2 + xy + y2 can never be negative if x and y are
integers can be proved because x2 + y2 is always positive,
and always bigger than xy. The fact that
x2 + xy + y2 = 0 only if both x = 0 and y = 0 if x and y
are integers can be proved using the discriminant. If x2 + xy + y2 = 0,
then the discriminant of this quadratic equation is b2 – 4ac = y2
– 4∙1∙y2 = -3y2.
If we assume that y ≠ 0, then -3y2 is always negative, which
would mean x has two imaginary solutions, which would contract that x is an
integer. So y = 0. By a similar argument (switching x and y), x
= 0.
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