Bunuel wrote:
If p is a prime number greater than 3, find the remainder when p^2 + 17 is divided by 12.
A. 8
B. 7
C. 6
D. 1
E. 0
Test any number. Let p = 5 for example. \(p^2 + 17 = 42\). 42 divided by 12 has a remainder of 6.
Ans: C
For the proof of why the remainder is always 6:
Let n be any prime number greater than 3. We can write n as \(n = E+ 1\)
where E is even.
Then \(n^2 = (E + 1)^2 = E^2 + 2E + 1\).
The claim is \(E^2 + 2E\) will always have a factor of 12. We need to prove it has a factor of both 4 and 3.
The factor of 4 is easier to prove. Since \(E^2 + 2E = E*(2+E)\), both E and (2+E) have a factor of 2 so the product must have a factor of 4.
For the factor of 3, recall E + 1 is a prime number greater than 3 so it cannot be divisible by 3. Thus either E or E + 2 must be divisible by 3. Both are factors of the product, therefore in any case \(E^2 + 2E\) must be divisible by 3.
Finally, we proved \(E^2 + 2E\) is divisible by 12, then \(n^2 = E^2 + 2E + 1\) divided by 12 will have a remainder of 1 and \(n^2 + 17\) must have a remainder of 6.
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