2
$\begingroup$

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

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

$\endgroup$
  • 10
    $\begingroup$ Since these questions are unrelated to each other, please post each one as a new question. $\endgroup$ – Clayton Mar 21 '13 at 4:37
5
$\begingroup$

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

$\endgroup$
4
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.