I'd approach it by first noting the statement is equivalent to stating that at least one of (n, n+2, n²+3) has a divisor p², where p is a prime; and then use an indirect proof:
Assume the statement is false. Then there is at least on value n such that none of (n, n+2, n²+3) has a square divisor. Then n can't be even, since that would make either n or n+2 divisible by 4.
So let n = 2k+1 for some integer k. Then n² + 3 = (2k + 1)² + 3 = 4k² + 4k + 4 is divisible 4.
That's impossible, so the assumption must be false.
That actually demonstrates the stronger statement that for any integer, one of (n, n+2, n²+3) is specifically divisible by 4.
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Verified answer
if n == 0 (mod 4) then µ(n) = 0
if n == 1 (mod 4) then n^2 + 3 == 4 == 0 (mod 4), so µ(n^2 + 3) = 0
if n == 2 (mod 4) then n + 2 == 0 (mod 4) so µ(n + 2) = 0
if n == 3 (mod 4) then n^2 + 3 == 3^2 + 3 = 12 == 0 (mod 4), so µ(n^2 + 3) = 0
In any case µ(n)µ(n+2)µ(n^2+3) = 0.
I'd approach it by first noting the statement is equivalent to stating that at least one of (n, n+2, n²+3) has a divisor p², where p is a prime; and then use an indirect proof:
Assume the statement is false. Then there is at least on value n such that none of (n, n+2, n²+3) has a square divisor. Then n can't be even, since that would make either n or n+2 divisible by 4.
So let n = 2k+1 for some integer k. Then n² + 3 = (2k + 1)² + 3 = 4k² + 4k + 4 is divisible 4.
That's impossible, so the assumption must be false.
That actually demonstrates the stronger statement that for any integer, one of (n, n+2, n²+3) is specifically divisible by 4.