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These are all questions I would never even have asked myself until that incident with Don. Every day, my friend Don and I would see who could trip each other the most times. But then one day I tripped him and he fell and broke his jaw. He looked up and, with slurred speech, said, “I guess you win.” But what did I win? I didn't win anything, and you know why? Because I forgot to make a bet with him. But something else was wrong, and I knew it. Why did I want to trip Don in the first place? To show how clever I was, or how brave, or how successful? Yes, all those things. So I guess that answers that.  Jack Handey
1d

comment 
On the number of consecutive divisors of an integer
Here's a PDF: renyi.hu/~p_erdos/197826.pdf 
1d

answered  On the number of consecutive divisors of an integer 
May 14 
answered  On the natural density of almost perfect numbers 
May 10 
revised 
Are There Infinitly Many $n$ Which $a\times n!+1$ Be Composite?
deleted extraneous plus sign 
May 9 
answered  Are There Infinitly Many $n$ Which $a\times n!+1$ Be Composite? 
May 4 
answered  Powers modulo a fixed integer 
Apr 18 
comment 
When has the BorelCantelli heuristic been wrong?
Another example of (probable) failure in the presence algebraic structure: Pomerance has a heuristic, described at oddperfect.org/pomerance.html, predicting no odd perfect numbers. The same heuristic could be adapted to suggest there are only finitely many even perfect numbers  but of course we "know better" here, by Euclid. 
Mar 26 
awarded  Nice Answer 
Mar 26 
comment 
Numbers $n$ such that the sum of the divisors of $n$ is a nontrivial power
Let me supplement my previous comment by saying that it's not realistic at this time to expect definitive results for fixed $a$. For example, consider the equation $\sigma(n) = 6^b$. Probably there are infinitely many solutions $n,b$, since if $p=2\cdot 6^m1$ is prime, then $n=2p$ gives rise to a solution. On the other hand, it seems hopeless to prove at this time that there are infinitely many solutions. For then there are infinitely many $p$ with $p+1$ a $3$smooth number. We don't know anything nearly this strong! The ``smoothest'' we can get $p+1$ is $p^{0.2931}$ (BakerHarman)! 
Mar 25 
answered  Does there exist an integer that is both solitary and almost perfect? 
Mar 25 
comment 
Numbers $n$ such that the sum of the divisors of $n$ is a nontrivial power
Here is a comment on the fixed $a$ problem: One can show that there are only finitely many proper prime powers $p^e$ for which $\sigma(p^e)$ is supported on a given finite set of primes. See, e.g., the book of Shorey and Tijdeman on Exponential Diophantine Equations. This means that the problem in some sense reduces to understanding the shifted primes $\sigma(p)=p+1$. For instance, $\sigma(n) = 2^b$ has infinitely many solutions $(n,b)$ if and only if there are infinitely many Mersenne primes. 
Mar 24 
answered  Numbers $n$ such that the sum of the divisors of $n$ is a nontrivial power 
Feb 11 
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Waring problem for binomial coefficients (generalization of Gauss' Eureka Theorem)
As Gjergji has pointed out, this is a consequence of a general theorem of Kamke. A modern reference is Nathanson's Elementary methods in Number Theory; see Chapter 11, especially Theorems 11.10 and 11.12. 
Feb 1 
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Surveys of the items of Erdős' “toolbox”
The recent volume that came out of the Erdős centennial conference might also be of interest to you: springer.com/new+%26+forthcoming+titles+%28default%29/book/… 
Dec 24 
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RuthAaron triples, etc
It seems that, contra Hoffman, Erdos never claimed to prove the infinitude of RuthAaron pairs. After the original RuthAaron paper appeared in the Journal of Recreational Mathematics, Pomerance learned of Erdos's interest from a letter (not a phone call). In the letter, Erdos says he can prove that the RuthAaron numbers have density $0$. Far from claiming a proof that there are infinitely many RuthAaron numbers, Erdos says that problem "seems hopeless" ! The original letter is available online  see the very first scan at cah.utexas.edu/collections/math_erdos.php 
Nov 24 
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Siegel Walfisz Theorem for algebraic number fields
An analogue of of SiegelWalfisz for Hecke characters can be found in the following paper: Goldstein, Larry Joel: A generalization of the SiegelWalfisz theorem. Trans. Amer. Math. Soc. 149 1970 417–429. 
Nov 10 
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A conjecture of Erdos on consecutive differences of primes
The problem is solved by recent work of BanksFreibergTurnageButterbaugh. Using the method of MaynardTao, they show that the sequence of e_k's contains arbitrarily long runs of 0's, as well as arbitrarily long runs of 1's. Hence, the sum has a nonperiodic binary expansion and so represents an irrational number. 
Oct 30 
awarded  Nice Answer 
Oct 25 
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Effective version of the BombieriVinogradov theorem
For effective variants of BombieriVinogradov, see e.g. this paper of Akbary and Hambrook: cs.uleth.ca/~akbary/AkbaryHambrook.pdf or Lemma 5.2 of this paper of Lenstra and Pomerance: math.dartmouth.edu/~carlp/PDF/complexity12.pdf 
Oct 25 
awarded  Nice Answer 