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| Are there any triples <math>(a,b,c)</math> of positive integers such that <math>(a-2)(b-2)(c-2) + 12</math> is prime that properly divides the positive number <math>a^2 + b^2 + c^2 + abc - 2017</math>? | | Are there any triples <math>(a,b,c)</math> of positive integers such that <math>(a-2)(b-2)(c-2) + 12</math> is prime that properly divides the positive number <math>a^2 + b^2 + c^2 + abc - 2017</math>? |
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− | ==Solution== | + | ==Solution 1== |
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− | Yes. Let <math>p = (a-2)(b-2)(c-2)+12 = abc - 2(ab+ac+bc)+4(a+b+c)+4</math>. Also define <math>\alpha=a+b+c</math>.
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− | We want <math>p</math> to divide the positive number <math>a^2+b^2+c^2+abc-2017=(\alpha-47)(\alpha+43)+p</math>. This equality can be verified by expanding the righthand side.
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− | Because <math>a^2+b^2+c^2+abc-2017</math> will be trivially positive if <math>(\alpha-47)(\alpha+43)</math> is non-negative, we can just assume that <math>\alpha=47</math>.
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− | Analyzing the structure of <math>p</math>, we see that <math>a-2</math>,<math>b-2</math>, and <math>c-2</math> must be <math>1</math> or <math>5</math> mod <math>6</math>, or <math>p</math> will not be prime (divisibility by <math>2</math> and <math>3</math>).
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− | Thus, we can guess any <math>a</math>,<math>b</math>, and <math>c</math> which satisfies those constraints.
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− | For example, <math>a = 21</math>,<math>b=13</math>, and <math>c=13</math> works. <math>p=2311</math> is prime, and it divides the positive number <math>a^2+b^2+c^2+abc-2017=2311</math>.
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− | This solution is wrong. No <math>p</math> actually exist.
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− | {{MAA Notice}}
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| ==See also== | | ==See also== |
| {{USAJMO newbox|year=2017|num-b=3|num-a=5}} | | {{USAJMO newbox|year=2017|num-b=3|num-a=5}} |