André Nicolas
Reputation
393/400 score
 2d comment How many distinct ways can the number be written as product of $3$ factors? Is the expression $2\times 63\times 1$ to be considered different from $1\times 2\times 63$? 2d comment For which primes $p\not=2$ is $5$ a square mod $p$? @wythagoras: There are marginally different definitions of quadratic residue modulo $p$. The standard textbooks I have used exclude $0$. That way, for odd primes $p$ there are exactly as many quadratic residues as non-residues, and the product of a residue and a non-residue is a non-residue. (That result would fail if we allow $0$ as a QR.) 2d comment For which primes $p\not=2$ is $5$ a square mod $p$? @wythagoras: I dealt separately with $5$ in the first paragraph. 2d answered For which primes $p\not=2$ is $5$ a square mod $p$? 2d comment Hilberts tenth problem over $\mathbb R$ with coefficients in $\mathbb Q$ @Halteproblem: Yes, that is correct. 2d comment For which primes $p\not=2$ is $5$ a square mod $p$? Can you use Quadratic Reciprocity? 2d comment Investigate convergence of $\sum_{n=1}^\infty \frac{\ln(n)}{n}$ For convergence/divergence, the first $2$ terms, or the first $20000$, do not matter. 2d revised Hilberts tenth problem over $\mathbb R$ with coefficients in $\mathbb Q$ added 138 characters in body 2d answered Hilberts tenth problem over $\mathbb R$ with coefficients in $\mathbb Q$ 2d comment Prove that $\sin ^{-1} 1$ is algebraic over $\mathbb Q$ Alternately, use the hint to show that $\cos\theta+i\sin\theta$ and $\cos\theta-i\sin\theta$ are both algebraic. Then use the fact that $\cos\theta$ is half the sum of these, and the sum of two algebraics is algebraic. The disadvantage of this method is that we have to use the non-trivial result that the sum of two algebraics is algebraic. Your suggested method is "closer to the ground." 2d comment Prove that $\sin ^{-1} 1$ is algebraic over $\mathbb Q$ Yes, your method for the second is correct. But you need to add some detail. When we expand, the real part gives $\cos^{17} \theta-\binom{17}{2}\cos^{15}\theta\sin^2\theta+\cdots=-1$. There will be various even powers of $\sin\theta$. Replace $\sin^2\theta$ everywhere by $1-\cos^2\theta$. 2d comment Show that $\cot \frac{\pi}{2m}\cot \frac{2\pi}{2m}\cot \frac{3\pi}{2m}…\cot \frac{(m-1)\pi}{2m}=1$ Yes, this one is very quick. It uses much less machinery that the $\cos$ one referred to in your question. 2d comment Prove that $\sin ^{-1} 1$ is algebraic over $\mathbb Q$ You probably know what $\sin^{-1} 1$ is (I assume that you mean $\arcsin 1$). For the second, think about $(\cos(\pi/17)\pm i\sin(\pi/17))^{17}$. 2d comment We have $a^3+b^3$ and $ab$, how we can calculate $a+b$? The general cubic is reducible to this form. So one cannot do better than the "Cardano" formula, or (if we are willing to use transcendental functions) the solutions of the cubic using trig/hyperbolic functions. 2d comment Show that $\cot \frac{\pi}{2m}\cot \frac{2\pi}{2m}\cot \frac{3\pi}{2m}…\cot \frac{(m-1)\pi}{2m}=1$ @GRS: You are welcome. Note that the basic idea is similar to the one for adding up $1+2+3+\cdots+50$. The sum of the first and last is $51$, as is the sum of second and next to last, and so on.In our case it is multiplication, not addition, but there is a similar "symmetry" thing going on. 2d comment Show that $\cot \frac{\pi}{2m}\cot \frac{2\pi}{2m}\cot \frac{3\pi}{2m}…\cot \frac{(m-1)\pi}{2m}=1$ You can also get it from $\sin(\pi/2-x)=\cos x$, $\cos(\pi/2-x)=\sin x$. 2d comment Show that $\cot \frac{\pi}{2m}\cot \frac{2\pi}{2m}\cot \frac{3\pi}{2m}…\cot \frac{(m-1)\pi}{2m}=1$ @GRS: You are welcome. You also know the formula, in the equivalent form $\tan(90^\circ-x)=\frac{1}{\tan x}$. Switch to the complementary angle and the $\tan$ flips. 2d comment Show that $\cot \frac{\pi}{2m}\cot \frac{2\pi}{2m}\cot \frac{3\pi}{2m}…\cot \frac{(m-1)\pi}{2m}=1$ @GRS: Odd number of terms (that is, $m$ even), works too, in that case the middle term is $\cot(\pi/4)$, which is $1$. 2d revised Show that $\cot \frac{\pi}{2m}\cot \frac{2\pi}{2m}\cot \frac{3\pi}{2m}…\cot \frac{(m-1)\pi}{2m}=1$ added 19 characters in body 2d comment The area of trapezium is given by $A=(a^2-x^2)(x+a)$. Find x for the area to be a maximum and find A max. You are welcome. For the maximum area plug into the formula for $A$. And $a$ is a number, albeit an unspecified one. The parabola has equation $y=a^2-x^2$, so it is not completely numerically specified.