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Then according to Lemma 2.1, there exists an odd integer and such that.
Then according to Lemma 2.1, there exists an odd integer and such that .
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If there exists an odd positive integer m satisfying σ 1 ( m ) = 2 m, then the number of n satisfying σ 1 ∗ ( n ) = 2 n is infinite, that is, σ 1 ∗ ( 2 l m ) = 2 l σ 1∗∗ ( m ) = 2 ( 2 l m ).
If there exists an odd positive integer m satisfying σ 1 ( m ) = 2 m, then the number of n satisfying σ 1 ∗ ( n ) = 2 n is infinite, that is, σ 1 ∗ ( 2 l m ) = 2 l σ 1∗∗ ( m ) = 2 ( 2 l m ). . (Sophie Germain primes) Note that if there are an infinite number of Sophie Germain primes, then the problem σ 1 ∗ ( n ) = σ 1 ∗ ( n + 1 ) has an infinite number of solutions (in terms of prime numbers).
For a closed symmetric set A that does not contain the origin, we define the genus (gamma(A)) of A as the smallest integer k such that there exists an odd continuous mapping from A to (mathbb{R}^{k}setminus{0}).
Then, for any, there exists an odd homeomorphism.
If there exists an odd continuous mapping from A to B, then γ ( A ) ≤ γ ( B ).
Take the same as in Theorem 3.1, then for any, there exists an odd homeomorphism.
(1) If there exists an odd continuous mapping from A to B, then (gamma(A) leq gamma(B)).
If for any integer, then there exists an integer such that.
for all nonnegative integers n or there exists a positive integer n 0 such that.
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Justyna Jupowicz-Kozak
CEO of Professional Science Editing for Scientists @ prosciediting.com