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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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| − | Let <math>p = (a-2)(b-2)(c-2) + 12</math> and <math>m = a^2 + b^2 + c^2 + abc - 2017</math>. We have that:
| + | ==Solution 1== |
| − | <cmath>m - p = a^2 + b^2 + c^2 + 2ab + 2ac + 2bc - 4a - 4b - 4c - 2021 = (a+b+c-2)^2 - 45^2</cmath>Let <math>a+b+c = x</math>. Then <math>m - p</math> is <math>(x-2)^2 - 45^2</math>, which must be divisible by <math>p</math>.
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| − | Since <math>m > p</math>, <math>x > 47</math>, and since <math>p</math> divides <math>m - p</math>, <math>p</math> must divide either <math>x-47</math> or <math>x+43</math>.
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| − | It is easy to see that the minimum of <math>p = (a-2)(b-2)(c-2)+12</math> is <math>x+4</math>. Since <math>p > x+4 > x-47</math>, <math>p</math> cannot divide <math>x-47</math>, so <math>p</math> must divide <math>x+43</math>. If <math>p \not= x+43</math>, <math>x+43 \ge 2p</math>. But <math>x + 43 < 2x + 8 < 2p</math>, so <math>p = x+43</math>. If <math>p</math> is prime (p > 47 + 43 = 90), then <math>x</math> has to be even, making one of <math>a,b,c</math> even, making <math>(a-2)(b-2)(c-2) + 12</math> an even number, which is a contradiction.
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| − | Thus, there are no integer triples that work.
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| − | ~AopsUser101
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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}} |
of positive integers such that 
is prime that properly divides the positive number 
?
