1.Find all positive integers solution

$xy+yz+xz = xyz+2$

2.Determine all p and q which p,q are prime number and satisfy

$p^3-q^5 = (p+q)^2$

Thx for the answer

3.Find all both positive or negative integers that satisfy

$\frac{13}{x^2} + \frac{1996}{y^2} = \frac{z}{1997}$

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    $\begingroup$ Since these questions are unrelated to each other, please post each one as a new question. $\endgroup$ – Clayton Mar 21 '13 at 4:37

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 $a=0, 1, 2, 3, 4, 5$.


1.set $x≤y≤z,xyz<xyz+2=xy+yz+zx≤3yz,$

so $0<x<3,x=1,2$

If $x=1,y+z+yz=yz+2,y+z=2$,so $x=y=z=1$

If $x=2,2y+2z+yz=2yz+2,yz=2y+2z-2<4z,2≤y<4,y=3,z=4$

so the only solution to the fisrt equation is $x=y=z=1$ or $x=2,y=3,z=4$ or change their order.


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