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Mediterranean Mathematical Competition

Algebra Difficulty 5.5 AIME, harder Prove it Spain

Determine all integers n≥1n \ge 1 for which the number n8+n6+n4+4n^8 + n^6 + n^4 + 4 is prime.

Solution

We have
n8+n6+n4+4=(n4−n3+n2−2n+2)(n4+n3+n2+2n+3)=f(n)g(n) n^8 + n^6 + n^4 + 4 = (n^4 - n^3 + n^2 - 2n + 2)(n^4 + n^3 + n^2 + 2n + 3) = f(n)g(n)
The first factor f(n)f(n) satisfies
f(n)=n4−n3+n2−2n+2=n3(n−1)+(n−1)2+1 f(n) = n^4 - n^3 + n^2 - 2n + 2 = n^3(n-1) + (n-1)^2 + 1
and hence f(n)≥2f(n) \ge 2 for all n≥2n \ge 2. The second factor g(n)=n4+n3+n2+2n+3g(n) = n^4 + n^3 + n^2 + 2n + 3 is strictly greater than 2 for all n≥2n \ge 2. This only leaves the case n=1n = 1 as a potential candidate for a prime, and indeed f(1)g(1)=1⋅7=7f(1)g(1) = 1 \cdot 7 = 7 is prime.

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