TITLE: Determine total turns $N_g$ before the coin slips from the rotating horisontal disk
QUESTION [0 upvotes]: Question: A coin with mass m is placed on a horisontal disk with distance $R$ from the centre of the disk. The disk is rotating at a constant acceleration $  \ddot \theta = a > 0$ and $a(t=0) = 0$ and velocity $v(t=0) = 0$ where $t \geq 0$. The disk is starting at rest. The static friction constant is $u_s$.
Determine the total amount of turns/laps $N_g$ before the coin starts slipping (A lap/turn is $2 \pi$ radians).
Attempt: The magnitude of the frictional force which keeps the coin from slipping outwards from the center is $F(t) \leq u_s \cdot mg$. Let $t=t_s$ be the time when the coin starts slipping. At the time of slipping we have maximum frictional force (1):
$Max[F(t)] = F(t_s) = u_s \cdot mg$.
For circular motion we have the acceleration $v^2/R = R \cdot \dot \theta^2$ towards the center of the disk.
Newtons second law then gives us (2):
$F(t_s) = m \cdot R \cdot \dot \theta^2$.
Solving equation (1) = (2) with respect to $\dot \theta^2$ gives us, at time $t=t_s$ (3):
$\dot \theta^2 = \frac{u_s \cdot g}{R}$.
Since $\ddot \theta = a$ and $\dot \theta (t=0) = 0$ we have $\dot \theta (t) + \dot \theta (t=0) = a \cdot t + 0 = a \cdot t = \dot \theta (t) $.
Now, with $a \cdot t = \dot \theta (t) $ we let $t=t_s$ and solve the following equation with respect to $t_s$:
$ (3) = \dot \theta^2 (t_s) = a^2 \cdot t_s^2 \leftrightarrow \frac{u_s \cdot g}{R} =  a^2 \cdot t_s^2 \leftrightarrow t_s = \sqrt{\frac{u_s \cdot g}{R \cdot a^2}} = (4)$.
We now have the time (4) when the coin starts slipping. Let total number of laps $N_g$ be a function of time $t: N_g (t)$ and can obviously be expressed as $\frac{\theta (t)}{2\cdot \pi}$. So in order to determine $N_g (t_s)$ we are looking for $\theta (t_s)$:
$\ddot \theta (t) = a$
$\dot \theta (t) = a \cdot t$
$ \theta (t) = \frac{a \cdot t^2}{2}$
$ \theta (t_s) = \frac{u_s \cdot g}{2 \cdot R \cdot a} = (5)$
With (5) we finally have the expression for $N_g (t_s)$ :
$N_g (t_s) = \frac{ \theta (t_s) }{2\cdot \pi} = \frac{u_s \cdot g}{4 \cdot \pi \cdot R \cdot a}$
Sadly, this is the wrong answer. Any help appreciated, especially if you can tell me where in my solution I do wrong. I think my error is that I did not include both frictional composants.

REPLY [0 votes]: Yes that's right - you have not included both frictional components, ie tangential as well as radial.
In the rotating frame of reference there are 2 fictitious forces acting on the stationary coin. One is the radial centrifugal force, which you have identified. This corresponds to radial acceleration $r\omega^2$ in the inertial frame of reference. The 2nd fictitious force is tangential, due to the angular acceleration $\alpha$ of the disk. This acts opposite to the direction of tangential acceleration in the inertial frame and causes tangential acceleration $r\alpha$ in the inertial frame of reference.
The real static friction force opposes the vector resultant of these two fictitious forces. You must take both into account in deciding when the static limit of friction has been reached.


TITLE: Finding formula that solves $w^4+x^4=y^4+z^4$ over the integers.
QUESTION [4 upvotes]: Several formulae that solve the diophantine equation
$$w^4 + x^4 = y^4+z^4 \tag{1}$$
are presented in this collection. The simplest one bases on
$$f_1 = a^7 + a^5 - 2 a^3 + 3 a^2 + a \tag{2}$$
and sets$\def\hf{{}^h\!f}$
$$ (w, x, y, z) = (\hf_1,\ \hf_1(b,-a),\ \hf_1(a,-b),\ \hf_1(b,a)) $$
where $\hf(a,b) = f(a/b)\cdot b^{\deg f}$ denotes the homogenized version of $f$.
Question: How does one find such formulae?
The collection refers to Hardy & Wright, and MathWorld mentions that parametric solutions to (1) were already known to Euler in 1802, but without mentioning which one.
Note: $f_1$ appears to be — in some sense — the simplest of such functions:  It produces (158, 59, 134, 133) from (1,2), the smallest solution of (1). All solutions that I checked using Sage were non-trivial except for the case $a=b=1$, and with $\gcd(a,b)=1$ all solutions satisfied $\gcd(w,x,y,z)\in\{1,2\}$.  The last two properties also showed up for the next 2 more complicted functions from the collection that I tried like
$$f_2 = -a^{13} + a^{12} + a^{11} + 5 a^{10} + 6 a^{9} - 12 a^{8} - 4 a^{7} + 7 a^{6} - 3 a^{5} - 3 a^{4} + 4 a^{3} + 2 a^{2} - a + 1
$$

REPLY [3 votes]: From L. E. Dickson, History of the Theory of Numbers, Volume II, Chapter XXII, page $644$

$$ A^4+B^4=C^4+D^4. $$ L. Euler$^{165}$ took
$\,A=p+q,\;\;D=p-q,\;\;C=r+s,\;\;B=r-s\,$
and derived
$$(1)\qquad\qquad\qquad pq(p^2+q^2)=rs(r^2+s^2).$$
Set $\,p=ax,\,q=by,\,r=kx,\,s=y.\,$ Then
$$ y^2/x^2=(k^3-a^3b)/(ab^3-k). $$
If $\,k=ab,\,x=1,\,$ then $\,y=\pm a,\,C=\pm A,\,B=\mp D.\,$
Set therefore $\,k=ab(1+z).\,$
Then $\, y^2/x^2=a^2Q/(b^2-1-z)^2,\,$ where
$$ Q\!=\!(b^2\!-\!1)^2\!+\!(b^2\!-\!1)(3b^2-1)z \!+\!
3b^2(b^2\!-\!2)z^2\!+\!b^2(b^2\!-\!4)z^3\!-\!b^2z^4. $$
Let $\,Q\,$ be the square of $\,b^2-1+fz+gz^2\,$ and
choose $\,f,g\,$ to make the terms in $\,z,z^2\,$ agree. Thus,
$$ f=\frac{3b^2-1}2,\quad g=\frac{3b^4-18b^2-1}{8(b^2-1)},\quad
z=\frac{b^2(b^2-4)-2fg}{b^2+g^2}. $$
Then $\,x:y=b^2-1-z:a(b^2-1+fz+gz^2).\,$
As examples, Euler took $\,b=2,\,b=3\,$ and found the solution
$$ A=2219449,\quad B=-555617,\quad C=1584749,\quad D=2061283,$$
and an erroneous$^{166}$ one replaced in his next paper by
$$ A=12231,\quad B=2903,\quad C=10381,\quad D=10203.$$

I have verified the steps for Euler as quoted in Dickson. For $\,b=1\,$
the result is $\,A=1,\,B=-1,\,C=1,\,D=-1\,$ and for $\,b=2,\,b=3\,$ the
results are as given.

The general result given by Euler depends on one parameter $\,b.\,$
Define the two variable polynomial
$$ P(u,v) := u^{13}+27 u^{12} v-214 u^{11} v^2-186 u^{10} v^3-
2481 u^9 v^4+861 u^8 v^5-2804 u^7 v^6\\-972 u^6 v^7-2481 u^5 v^8-
27 u^4 v^9-214 u^3 v^{10}+294 u^2 v^{11}+u v^{12}+3 v^{13}. $$
The solution given by Euler is essentially
$$ A=P(-b,1),\quad B=P(1,b),\quad C=-P(1,-b),\quad D=P(b,1).$$
Note that $\,P(u,v) = {}^h\!f_2(u+v,v-u)\,$ is the connection with
your solution $\,f_2.$
The polynomial corresponding to your $\,f_1\,$ is
$$ P(u,v) := u^6 v+3 u^5 v^2-2 u^4 v^3+u^2 v^5+v^7 $$
due to Gerardin which I got from A Collection of Algebraic Identities by Tito Piezas III.
Dickson mentions Gerardin briefly on page 647

A. Gerardin$^{181a}$ noted that $(1)$ has the solution
$$ p=a^7+a^5-2a^3+a,\quad q=3a^2,\quad r=a^6-2a^4+a^2+1,\quad s=3a^5,$$
which is simpler than Euler's solution $(5)$.

Your question

How does one find such formulae?

is a difficult one. Another answer mentions
$K3$ surfaces
of which I know almost nothing. Particular cases of interest
include $\,A^4+B^4+C^4+D^4=0\,$ which is the Fermat quartic.
The article
Euler's sum of powers conjecture
mentions the case $\,A^4+B^4+C^4=D^4\,$ which was solved by
Noam Elkies using elliptic curves.