Since @Hecke has done the first one, I shall only provide the solutions for 2 and 3.
2. $p^5-q^5>p^3-q^5=(p+q)^2 \geq 0$ so $p>q$. Thus $\gcd(p, q)=\gcd(q, p+q)=\gcd(p, p+q)=1$.
Taking $\pmod{p}$, we get $p \mid q^5+q^2=q^2(q+1)(q^2-q+1)$, so $p \mid (q+1)$ or $p \mid (q^2-q+1)$.
If $p \mid q+1$, then since $p>q$, we have $p=q+1$. Now $$0<(2q+1)^2=(q+1)^3-q^5 \leq (\frac{3}{2}q)^3-q^5=q^3(\frac{27}{8}-q^2)<0$$
since $q^2 \geq 4>\frac{27}{8}$. We thus get a contradiction.
Therefore, $p \mid (q^2-q+1)$, and thus $p \leq q^2-q+1$.
Taking $\pmod{p+q}$, we get $(p+q) \mid -q^3-q^5=-q^3(1+q^2)$. Since $\gcd(q, p+q)=1$, we get $(p+q) \mid (q^2+1)$.
Now $p \mid ((q^2-q+1)-p), (p+q) \mid ((q^2-q+1)-p)$, so since $\gcd(p, p+q)=1$, $p(p+q) \mid ((q^2-q+1)-p)$.
If $p<q^2-q+1$, then we must have $((q^2-q+1)-p) \geq p(p+q)$.
But then $p^2+1>q^2+1 \geq (p+1)(p+q)>(p+1)p$, a contradiction.
Thus $p=q^2-q+1$, and we get $(q^2-q+1)^3-q^5=(q^2+1)^2$.
Now taking $\pmod{q-1}$, we get $0 \equiv 1-1 \equiv (q^2-q+1)^3-q^5 \equiv (q^2+1)^2 \equiv 2^2 \pmod{q-1}$
Thus $(q-1) \mid 4$, and so $q=2, 3, 5$.
If $q=2$, then $-5=3^3-2^5=5^2=25$, a contradiction.
If $q=3$, then $100=7^3-3^5=10^2=100$, and we get a solution $(p, q)=(7, 3)$.
If $q=5$, then $6136=9261-3125=21^3-5^5=26^2=676$, a contradiction.
Thus the only solution is $(p, q)=(7, 3)$.
3. I don't really understand the meaning of "all both positive or negative integers". Do you mean $x, y, z$ are all positive integers or all negative integers, or do you mean that $x, y, z$ are all integers, and $x, y$ are both positive or both negative? If it is the former, note that $z>0$, so then $x, y>0$. If it is the latter, which in my opinion seems more likely, better phrasing is needed.
$\frac{13}{x^2}+\frac{1996}{y^2}=\frac{z}{1997}$.
Let $\gcd(x, y)=d, d>0, x=dx', y=dy'$, then $1997(13y'^2+1996x'^2)=zd^2x'^2y'^2$
Now $x'^2 \mid 1997(13)y'^2$ so since $\gcd(x', y')=1$, $x'^2 \mid 1997(13)$. $(1997)(13)$ is squarefree, so $x'=\pm 1$. Similarly, $y'^2 \mid (1997)(1996)x'^2$, so $y'^2 \mid (1997)(1996)=1997(499)(2^2)$. $1997(499)$ is squarefree, so $y'= \pm 1, \pm 2$
If $y'=\pm 1$, then $zd^2=1997(13+1996)=1997(41)(7^2)$, so $d \mid 7$.
If $d=1$, we get $z=1997(41)(7^2)=4011973$. We thus get the solutions $(x, y, z)=(\pm 1, \pm 1, 4011973)$.
If $d=7$, we get $z=1997(41)=81877$. We thus get $(x, y, z)=(\pm 7, \pm 7, 81877)$.
If $y'=\pm 2$, then $zd^2=1997(13 \cdot 4+1996)=1997(2048)=1997(2^{11})$. Thus $d \mid 2^5$.
Let $d=2^a, a=0, 1, 2, 3, 4, 5$, then $z=1997(2^{11-2a})$, and we get $(x, y, z)=(\pm 2^a, \pm 2^{a+1}, 1997(2^{11-2a}))$.
In conclusion, the solutions are $(x, y, z)=(\pm 1, \pm 1, 4011973), (\pm 7, \pm 7, 81877)$, or $(\pm 2^a, \pm 2^{a+1}, 1997(2^{11-2a}))$, where $a=0, 1, 2, 3, 4, 5$.
Try these monsters:
$x_{n_1}=\frac{1}{2a}\left[\frac{\gamma_0 \sqrt{P^2-1}+Q\beta_0}{2\sqrt{P^2-1}}\cdot \left(P+\sqrt{P^2-1}\right)^n+ \frac{\gamma_0 \sqrt{P^2-1}-Q\beta_0}{2\sqrt{P^2-1}}\cdot \left(P-\sqrt{P^2-1}\right)^n-b\right]$
and
$x_{n_2}=\frac{1}{2a}\left[\frac{\gamma_0 \sqrt{P^2-1}-Q\beta_0}{2\sqrt{P^2-1}}\cdot \left(P+\sqrt{P^2-1}\right)^n+ \frac{\gamma_0 \sqrt{P^2-1}+Q\beta_0}{2\sqrt{P^2-1}}\cdot \left(P-\sqrt{P^2-1}\right)^n-b\right]$
6 notes:
- $(a,b,c)$ from $f(x)=ax^2+bx+c$
- use minimal $(P,Q)$ such that $P^2-aQ^2=1$
- use minimal $(\gamma_0,\beta_0)$ such that $a\gamma_0^2-\beta_0^2=a(b^2-4ac)$
- for non-square $a$, when a is a square it's analogous to the explanation below
- Reasoning for $x_{n_1}$ and $x_{n_2}$ is that some solutions to $ \ a\gamma^2-\beta^2=a(b^2-4ac) \ $ have two separate strands.
- In using this as written, it may be that you need to take every $ \ n=2m \ $ or $ \ n=2m-1 \ $ or multiples $n=km$ to meet the $\gamma-b \equiv 0 \pmod {2a}$ constraint (referring to how $ \ 2a \cdot x + b = \gamma \ $ below)
The derivation of the above is arriving at the pell equation:
$$\begin{align} ax^2 + bx + c &= f(x)=\alpha^2 \\
a(x+b/2a)^2-\frac{b^2-4ac}{4a} &=\alpha^2 \\
a(2ax+b)^2-a(b^2-4ac)&=4a^2\alpha^2=\beta^2 \\
a\gamma^2-\beta^2&=a(b^2-4ac)
\end{align}$$
And solving using standard techniques.Thus
$$x_n=\frac{\gamma_n-b}{2a}$$
Such that $\gamma_n$ is the nth solution in the above pell type equation : $a\gamma^2-\beta^2=a(b^2-4ac)$
An explanation as to why in your first problem $(2,10,19,47)$ are the only answers is what follows:
The problem at the top has $(a,b,c)=(4,84,-15)$. Since $a$ is square, that pell equation morphs into a difference of two squares.
$$ \begin{align}
4\gamma^2-\beta^2&=29184 \\
(2\gamma)^2-\beta^2&=29184=2^9\cdot3\cdot19
\end{align}$$
Using this identity:
$$\left(\frac{d_1+d_2}{2}\right)^2-\left(\frac{d_1-d_2}{2}\right)^2=d_1\cdot d_2$$
Let $d_1, d_2$ be two divisors (of the same parity) of $29184$. You can see here that since there can only be a limited number of divisors of $29184$, even more so limited that have the same parity, you know right here that there's going to be a limited number of solutions.
Let $2\gamma=\frac{d_1+d_2}{2}$, thus $\gamma=\frac{d_1+d_2}{4}$, and finally putting this back into our $x$, and letting $b=84$ and $a=4$, arrive at
$$x=\frac{\left(\frac{d_1+d_2}{4}\right)-84}{8}$$
or
$$x=\frac{d_1+d_2-336}{32}=\frac{d_1+d_2}{32}-\frac{21}{2}$$
Now we observe that both factors must add to something directly proportional to 16:
$$d_1+d_2=16\cdot k$$
for the fraction $21/2$ to become whole when added. This implies that both $d_1$ and $d_2$ need to have a factor of atleast $2^4$. The only two-factor sets that come from $2^9\cdot3\cdot19$ and meet this requirement are:
$$\begin{align}
(d_1,d_2) &\to x(d_1,d_2) &&\to x=\frac{d_1+d_2}{32}-\frac{21}{2} \\
\cdots \cdots \cdots \cdots \cdots \cdots \\
(3\cdot 19 \cdot 2^4, 2^5) &\to x(912,32) &&\to x=19 \\
(3\cdot 19 \cdot 2^5, 2^4) &\to x(1824,16) &&\to x=47 \\
(3\cdot 2^4,19 \cdot 2^5) &\to x(48,608) &&\to x=10 \\
(3\cdot 2^5,19 \cdot 2^4) &\to x(96,304) &&\to x=2 \\
\end{align}$$
Best Answer
If all the integers are greater than or equal to $2$, then $$ a+b+c < abc = x+y+z < xyz = a+b+c. $$ Contradiction. Therefore, assume WLOG that $z=1$. You get $$ a+b+c = xy, \qquad x+y+1 = abc. $$ Now assume again, $a, b, c, x, y\geq2$, you get $$ a+b+c < abc = x+y+1 \leq xy + 1 = a+b+c+1. $$ Therefore, $abc=a+b+c+1$ and $x+y=xy$. Since $x, y\geq 2$, this implies $x=y=2$. Easy to find $a, b, c$ from here (if there are any). The remaining cases are when $c=1$ or $y=1$.
$(a, b, c) = (3, 2, 1)$ and $(x, y, z) = (3, 2, 1)$ is one example.