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A question on residues mod an even integer
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On permuted sum of squares of primes in a list
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On permuted sum of squares of primes in a list
@JAS 6N by p_{31} to p_{36} and 5N by 37 to 41. This is possible by choosing $N$ to be large and $1\pmod {24}$ and using Hua's theorem. One also needs lots of ways of writing numbers as sums of five squares of primes. This is standard from the circle method, but I don't know a reference. See my comment to fedja also. |
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On permuted sum of squares of primes in a list
@JAS Here's an example if that helps of the proof. Suppose $k=41=30+6+5$. Then take the permutation $(1,2,...,30)(31,...,36)(37,...41)$ and its powers. These give $30$ permutations of the primes $p_1$ to $p_{41}$. If you add the vectors, the first thirty entries will match and they are the number $30N$ written as a sum of $30$ squares of distinct primes (assume all primes $\ge 5$ so the squares are $1\pmod{24}$). Next six entries will be $5$ times a number that is a sum of six squares; and the last $5$ entries will be $6$ times a sum of $5$ squares.This is why we also want $5N$ and $6N$. |
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On permuted sum of squares of primes in a list
@fedja Please note that Hua's theorem takes $n$ to be $5 \pmod {24}$ and that every prime square apart from $4$ and $9$ is $1 \pmod {24}$. As for the number of representations, this is why I wrote that a reference would be hard to find, but the number of representations will be very many --- about $n^{3/2+o(1)}$ for five squares of distinct primes. |
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On permuted sum of squares of primes in a list
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Sep 29 |
answered | On permuted sum of squares of primes in a list |
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Backlund counting formula for Dirichlet L-functions?
Did you look at Chapter 16 of Davenport's book on multiplicative number theory? The details are not given but the argument is standard. |
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A question on residues mod an even integer
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A question on residues mod an even integer
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A question on residues mod an even integer
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A question on residues mod an even integer
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A question on residues mod an even integer
You're right. So the structure theorem I referred to gives here that the odd numbers 1, 3, 11 are three numbers out of the progression 11, 13, 1, 3; and the even numbers 0, 2, 6 are three out of the progression 0, 2, 4, 6. So a result of that type is perhaps the best you could get. |
Sep 28 |
answered | A question on residues mod an even integer |
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How $a+b$ can grow when $a!b! \mid n!$
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How $a+b$ can grow when $a!b! \mid n!$
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