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| | <math>\{ 1,25,49,73,97 \}</math> | | <math>\{ 1,25,49,73,97 \}</math> |
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| − | If <math>{a^3}+23</math> is divisible by <math>24</math>, then so is <math>{a^3}-1</math>. <math>{a^3}-1</math> is the same as <math>(a-1)</math><math>({a^2}+a+1)</math>, and it is divisible by <math>24</math>. There are many options: <math>(a-1)</math> is divisible by <math>24</math>, <math>({a^2}+a+1)</math> is divisible by <math>24</math>, <math>(a-1)</math> is divisible by <math>8</math> and <math>({a^2}+a+1)</math> is divisible by <math>3</math>, and <math>(a-1)</math> is divisible by <math>3</math> and <math>({a^2}+a+1)</math> is divisible by <math>8</math>.
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| − | Case 1: <math>(a-1)</math> is divisible by <math>24</math>
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| − | This means that <math>a</math> is congruent to <math>1</math> <math>mod</math> <math>(24)</math>. Satisfying the range, the following integers that satisfy are:
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| − | {<math>1,25,49,73,97</math>}
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| − | Case 2: <math>({a^2}+a+1)</math> is divisible by <math>24</math>
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| − | Or <math>a(a+1)</math> is in the form <math>23</math> <math>mod</math> <math>(24)</math>. This means that <math>a(a+1)</math> can be {<math>23,47,71,95</math>}, in the list, the first three numbers are prime, and The fourth can be factorized into two non-consecutive primes. No results from this case.
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| − | Case 3: <math>(a-1)</math> is divisible by <math>8</math> and <math>({a^2}+a+1)</math> is divisible by <math>3</math>
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| − | <math>a</math> is congruent to <math>1</math> <math>mod</math> <math>(8)</math> or <math>a</math> = <math>8</math> <math>k</math> + <math>1</math> where <math>k</math> is a positive integer. This means that <math>{(8k+1)^2}</math> + <math>8k+1</math> + <math>1</math> = <math>64</math> <math>{k^2}</math> + <math>24</math> <math>k</math> + <math>3</math> which has to be divisible by <math>3</math>. That means so does <math>{k^2}</math>, or <math>k</math> itself is divisible by <math>3</math>. The maximum it can be <math>12</math>, because <math>a<100</math> or <math>a</math> = <math>8k+1</math>. However, for the available values that can be inputted (0,3,6,9,and 12),the same list results from Case 1. No new values.
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| − | Case 4: <math>(a-1)</math> is divisible by <math>3</math> and <math>({a^2}+a+1)</math> is divisible by <math>8</math>
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| − | <math>a</math> is congruent to <math>1</math> <math>mod</math> <math>(3)</math> or <math>a</math> = <math>3</math> <math>k</math> + <math>1</math> where <math>k</math> is a positive integer. This means that <math>{(3k+1)^2}</math> + <math>3k+1</math> + <math>1</math> = <math>9</math> <math>{k^2}</math> + <math>9</math> <math>k</math> + <math>3</math> which has to be divisible by <math>8</math>. That means so does <math>{k^2}</math> + <math>k</math> + <math>3</math>. Checking k modulo eight for all values might result in a value of k which can narrow down search values.
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| − | Sub case 1: k is congruent to -4(mod 8)
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| − | <math>{(-4)^2}</math> + <math>-4</math> + <math>3</math> = 7(mod 8)
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| − | Sub case 2: k is congruent to -3(mod 8)
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| − | <math>{(-3)^2}</math> + <math>-3</math> + <math>3</math> = 1(mod 8)
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| − | Sub case 3: k is congruent to -2(mod 8)
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| − | <math>{(-2)^2}</math> + <math>-2</math> + <math>3</math> = 5(mod 8)
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| − | Sub case 4: k is congruent to -1(mod 8)
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| − | <math>{(-1)^2}</math> + <math>-1</math> + <math>3</math> = 3(mod 8)
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| − | Sub case 5: k is congruent to 0(mod 8)
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| − | <math>{(-0)^2}</math> + <math>-0</math> + <math>3</math> = 3(mod 8)
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| − | Sub case 6: k is congruent to 1(mod 8)
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| − | <math>{(1)^2}</math> + <math>1</math> + <math>3</math> = 5(mod 8)
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| − | Sub case 7: k is congruent to 2(mod 8)
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| − | <math>{(2)^2}</math> + <math>2</math> + <math>3</math> = 1(mod 8)
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| − | Sub case 8: k is congruent to 3(mod 8)
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| − | <math>{(3)^2}</math> + <math>3</math> + <math>3</math> = 7(mod 8)
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| − | In no scenario is <math>{k^2}</math> + <math>k</math> + <math>3</math> divisible by <math>8</math>, and by working backward, neither can <math>({a^2}+a+1)</math>. This means that the list noted in Case 1 are all the numbers possible that satisfy the condition. Our answer is <math>\boxed{\textbf{(1,25,49,73,97)}}</math>
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| | == See also == | | == See also == |
such that 
and 
is divisible by 
.
