Primetastic
Number Theory
Level
4
What is the greatest common factor of all integers of the form where p is a prime number greater than 5?
The answer is 240.
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1 solution
Solution: Let f(p) = p^4 − 1 = (p − 1)(p + 1)(p^2 + 1). Note that f(7) = 2^5· 3 · 5^2 and f(11) = 2^4· 3 · 5 · 61.
We now show that their greatest common factor, 2^4· 3 · 5, is actually the greatest common factor of all numbers p^4 − 1 so described.
• Since p is odd, then p^2 + 1 is even. Both p − 1 and p + 1 are even, and since they are consecutive even integers, one is actually divisible by 4. Thus, f(p) is always divisible by 2^4 . • When divided by 3, p has remainder either 1 or 2. – If p ≡ 1, then 3|p − 1. – If p ≡ 2, then 3|p + 1. Thus, f(p) is always divisible by 3. • When divided by 5, p has remainder 1, 2, 3 or 4. – If p ≡ 1, then 5|p − 1. – If p ≡ 2, then p^2 + 1 ≡ 2^2 + 1 = 5 ≡ 0. – If p ≡ 3, then p^2 + 1 ≡ 3^2 + 1 = 10 ≡ 0. – If p ≡ 4, then 5|p + 1.
Thus, f(p) is always divisible by 5.
Therefore, the greatest common factor is 2^4· 3 · 5 = 240