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45m
comment Are there rings whose multiplicative identity is not the number 1 or number 1-based?
Similar examples abound. Using an idempotent of a bigger ring as the neutral element is the common idea.
2h
comment How to create very hard problems on Lagrange Multipliers
Have you searched the site? Running the risk of excessive self-promotion I rather liked this trick. Not sure whether I would call it very hard though.
3h
comment Making change with prime-valued coins
Hint for part b. Have you ever heard of Goldbach's conjecture? Because it has been verified up to gazillions (with the expanded mint selection of part d), it seems clear that the answer won't be a small ____ number. It looks like you need to roll up your sleeves and do some dirty work listing sums that can be paid using at most two coins, and check what's the smallest one missing. Of course, you can try and be clever, but since this is contest math, I won't spoil this any further.
3h
answered Are there rings whose multiplicative identity is not the number 1 or number 1-based?
3h
comment Making change with prime-valued coins
Hint for part d. Assume that instead of separation of 6E we were asked about sequences with separation 2E. We see that 3,5,7 is such a sequence, but cannot find longer ones. In fact 3,5,7 is the only sequence of length 3, because if $n>2$, then not all of $n,n+2,n+4$ can be primes because one of them will be divisible by ___ (you fill in the blank and adapt).
4h
revised Isomorphism of sheaves
deleted 1 character in body
1d
awarded  Guru
1d
comment Solve $ 1^2+2^2+3^2+\cdots +k^2=n^2 $ in $ \mathbb{Z}^+ $
I suspect that this question may already have been answered here. Anyway, we would like to see a little bit more background and your ideas. Also, as $k$, $k+1$ and $2k+1$ are pairwise coprime, they would all need to be perfect squares apart from the factor $6$ distributed between them somehow. Looks like a tall order. And also a lot of case-by-case studies. May be a simpler method is out there :-)
1d
comment How many other hands of 13 cards containing 4 of any one suit and 3 of each of the other suits
There are many websites collecting information about Bridge related probabilities. Try here or here. Obviously I haven't checked all the data there, but generally the authors seem to know what they are doing :-)
1d
comment Composition of polynomials over finite fields
The power series of the form $x+\sum_{j>n}c_jx^j$ form a normal subgroup, call it $H_n$, of the Nottingham group $N(k_q)$. What you have is the quotient $N(k_q)/H_n$.
1d
comment Compute the splitting field and the Galois group of $x^4 - 5$ over $\mathbb{Q} (\sqrt{5})$.
Basically you are adjoining two square roots to $K=\Bbb{Q}(\sqrt{5})$. As you get a quartic extension of $K$, the square roots are "unrelated". So for the purposes of finding the Galois group the situation is analogous to finding the Galois group of, say, $\Bbb{Q}(\sqrt2,\sqrt3)/\Bbb{Q}$.
1d
comment Prove that $f=x^4-4x^2+16\in\mathbb{Q}$ is irreducible
@Lucas: The logic at that point was that because ithe polynomial has no real roots, it does not have any rational roots either. Therefore its has no linear factors, and we are left to handle the possibility of two quadratic factors. Sorry about not making that clear.
1d
comment Prove that $f=x^4-4x^2+16\in\mathbb{Q}$ is irreducible
+1 for this is a correct solution. The method is available whenever we can find the zeros. It is easier to come up with than the method I used. The downside is that this method is not available, if we cannot find the zeros.
1d
comment Prove that $f=x^4-4x^2+16\in\mathbb{Q}$ is irreducible
I though about doing it your way, @user17762. But I think this is simpler. Tastes vary :-) Both our techniques are important for newbies to master.
1d
answered Prove that $f=x^4-4x^2+16\in\mathbb{Q}$ is irreducible
1d
comment Find a counterexample to the following lemma if we change the statement slightly.
Sorry about some confusion initially. I need to rush. Hope it's correct now.
1d
revised Find a counterexample to the following lemma if we change the statement slightly.
added 45 characters in body
1d
answered Find a counterexample to the following lemma if we change the statement slightly.
1d
comment Is every Pisot-like integer the product of a Pisot integer and a root of unity?
Correct, @A.P. And thanks for spotting the error!
2d
answered Is every Pisot-like integer the product of a Pisot integer and a root of unity?