Squarefree quartics and power-free polynomial values

The explanation

Feed whole numbers into the formula n⁴ + 2. The input 1 gives 3. The input 2 gives 18, which is 2 × 3 × 3. The input 7 gives 2403, which is 3 × 3 × 3 × 89. Call a number squarefree when no prime appears twice in it. So 3 is squarefree, while 18 and 2403 are not. Among the inputs 1 to 5,000, exactly 3,787 give squarefree values. That is about 76%.

Where could a share like 76% come from? Check one prime at a time. The square of 2, which is 4, never divides n⁴ + 2. The square of 3, which is 9, divides it for 2 of the 9 possible remainders of n. So the prime 3 removes 2/9 of inputs and leaves 7/9. Each prime gives a survival fraction like this. Multiply them all together, and you get what is called an Euler product. For n⁴ + 2 it comes to about 0.757. The natural guess is that this product is the true long-run share.

The guess can fail in only one obvious way. Some prime square might divide every value of the formula. Then no value is squarefree at all. Call the formula admissible when that does not happen. The paper also needs the formula to be irreducibleSource 1. That means it cannot be split into smaller whole-number formulas multiplied together.

Paul Erdős singled out n⁴ + 2 as an open case in 1953. The general question allows any degree d, the highest power of n in the formula. It asks for values free of every prime raised to the power k. Such values are called k-power-free, so squarefree means 2-power-free. Ricci settled every k at least d in 1933. For k = d − 1, Erdős proved there are infinitely many such values. Hooley found their exact share in 1967.

The next step, k = d − 2, resisted. For degree 4 it is exactly the squarefree question. Nair's 1976 work reached k = d − 2 for degrees 24 and up. Heath-Brown's 2006 method brought this down to degree 10. Browning reached degree 9 in 2011. Degrees 4 to 8 stayed outside these ranges. Some other preprints claim special cases such as n⁴ + 1. This paper does not use those claimsSource 2.

This paper proves the guess for every admissible irreducible formula of degree 4 to 8Source 1. Here k = d − 2. A Lean proof is listed. For degree 4 that is the squarefree statement. For degrees 5, 6, 7 and 8, the excluded powers are cubes, fourth, fifth and sixth powersSource 3. Combined with Browning's theorem, this covers every degree from 4 upwardSource 4.

The hard part is large primesSource 5. Small primes are easy to handle by counting remainders, as with 3 above. But a value near n⁴ might be divisible by the square of a prime bigger than n itself. Counting remainders gives no control there. The paper shows that such inputs are rare enough to ignore. It turns each one into a point on a fixed surface, then shows such points are scarceSource 6.

So for n⁴ + 2, about 76% of all inputs give squarefree values, not just the first 5,000. The theorem gives this share as a limit. It does not say which inputs work, and it gives no rate for how fast the count settlesSource 7. The inputs are positive whole numbers, and the formula is fixed in advanceSource 8.

How sure is this? A computer proof checker called Lean has checked a formal version of the result. The Lean scope note states it for every degree from 4 upward under the admissibility condition. No specialists have published a review yet.

What it doesn't show. This does not say which inputs give squarefree values, or how fast the count nears its limit.