Proof. Let n ≥ 33 + 12 be an odd number. Since gcd(φ(32), 3) = 1, a runs through all odd residue classes modulo 32 as a runs through all odd residues modulo 32. Let P ∈ {3, 5, . . . , 33} be a cube such that P ≡ n − 12 mod 32. Then n − P = 4(8k + 3). Gauss showed that every positive integer 6≡ 0, 4, 7 mod 8 is a sum of three coprime squares [1, p. 262]. Hence, we may choose integers x, y, z with x + y + z = 8k + 3. Looking mod 8 we see x, y, z are all odd, so we may assume x, y, z ≥ 1. But then n = P + (2x) + (2y) + (2z). Therefore, for all odd n ≥ 33 + 12 = 35949, we have a representation of n as a sum of four proper powers. Let n ≥ 49 + 45 = 117693 be even. Since gcd(3, φ(48)) = 1, as a runs through all odd residues modulo 48, a also runs through all the odd residues modulo 48. Therefore, we can find an odd cube P ∈ {3, 5, . . . , 49} such that P ≡ n − 45 mod 48. Hence n = P + (48k + 45) for some integer k. Since 48k + 45 ≡ 5 mod 8, there exist integers x, y, z ≥ 0 such that x + y + z = 48k + 45 with gcd(x, y, z) = 1. We want to show none of x, y, z is 0 or 1. First, note that x + y ≡ 0 mod 3 implies x ≡ y ≡ 0 mod 3. If also z = 0, then gcd(x, y, z) ≥ 3. Therefore x, y, z ≥ 1. If one of x, y, z is 1, then (after relabeling) x + y = 48k + 44 ≡ 12 mod 16. The squares modulo 16 are 0, 1, 4, 9. We can’t add two of them to get 12 mod 16, therefore the representation of 48k + 45 as x + y + z includes only proper powers. Hence, n = P + x + y + z is the sum of four proper powers. Therefore, for n ≥ 49 + 45 we have a representation of n as a sum of four proper powers. To conclude one needs to check n < 117693. We computed all possible sums of 4 proper powers up to 117693 and we found all numbers between 1 and 117693 which are not represented as a sum of four proper powers.

中文翻译：

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适当权力的总和

证明。令 n ≥ 33 + 12 为奇数。由于 gcd(φ(32), 3) = 1，a 以模 32 的形式遍历所有奇数残基类别，因为 a 以模 32 的形式遍历所有奇数残基类。令 P ∈ {3, 5, ... . . , 33} 是一个立方体，使得 P ≡ n − 12 mod 32。那么 n − P = 4(8k + 3)。高斯证明了每个正整数 6≡ 0, 4, 7 mod 8 都是三个互质平方的和 [1, p. 262]。因此，我们可以选择整数 x、y、z，其中 x + y + z = 8k + 3。看模 8，我们看到 x、y、z 都是奇数，所以我们可以假设 x、y、z ≥ 1。但是然后n = P + (2x) + (2y) + (2z)。因此，对于所有奇数 n ≥ 33 + 12 = 35949，我们将 n 表示为四个真幂之和。令 n ≥ 49 + 45 = 117693 为偶数。由于 gcd(3, φ(48)) = 1，当 a 遍历所有以 48 为模的奇数余数时，a 也遍历所有以 48 为模的奇数余数。因此，我们可以找到一个奇数立方体 P ∈ {3, 5, ... . . , 49} 使得 P ≡ n − 45 mod 48。因此对于某个整数 k，n = P + (48k + 45)。由于 48k + 45 ≡ 5 mod 8，存在整数 x, y, z ≥ 0 使得 x + y + z = 48k + 45 且 gcd(x, y, z) = 1。我们不想显示 x， y, z 是 0 或 1。首先，注意 x + y ≡ 0 mod 3 意味着 x ≡ y ≡ 0 mod 3。如果 z = 0，则 gcd(x, y, z) ≥ 3。因此 x, y , z ≥ 1. 如果 x, y, z 之一是 1, 那么 (重新标记后) x + y = 48k + 44 ≡ 12 mod 16. 模 16 的平方是 0, 1, 4, 9. 我们不能将其中两个相加得到 12 mod 16，因此将 48k + 45 表示为 x + y + z 仅包括适当的幂。因此，n = P + x + y + z 是四个真幂之和。因此，对于 n ≥ 49 + 45，我们将 n 表示为四个真幂之和。结论是需要检查 n < 117693。