Verified answer

Given a (positive) integer n, let's write its prime factorization

n = p₁^(r₁) * p₂^(r₂) * ... * p_j^(r_j).

Then, τ(n) = (r₁ + 1) * (r₂ + 2) * ... * (r_j + 1)

and σ(n) = (1 + p₁ + ... + p₁^(r₁)) * ... * (1 + p_j + ... + p_j^(r_j)).

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Fix a positive integer k > 1.

1) Note that τ(n) = k has infinitely many solutions for n, because

τ(p^(k-1)) = k and there are infinitely many primes p.

(The tip-off for this question is that the primes in the factorization of n is irrelevant;

only their exponents matter for τ.)

2) On the other hand, σ(n) = k has at most finitely many solutions, because

σ(n) > n for all n > 1 and hence there are no more than k possible values of n (though in practice, much less) for which σ(n) = k.

I hope this helps!

If a+j=p_1^a_1*p_2^a_2*...*p_r^a_r is the best factorization of a+j, with each and every a_i>0, then a+j could have precisely (a_1+a million)*(a_2+a million)*...*(a_r+a million) divisors, which contains a million and a+j. for this reason interior the present concern, we ought to have (a_1+a million)*(a_2+a million)*...*(a_r+a million) =6. this suggests r<=2. In case r=a million, we ought to have a_1=5 and subsequently a+j=p^5 for some best p. In case r=2, (a_1,a_2)=(a million, 2) or (2,a million), so for this reason we ought to have a+j=p*q^2 for some primes p and q. subsequently the series a+a million, a+2, .., a+N ought to incorporate only integers that could nicely be factored into between the types p^5 or p*q^2.