In each of the following three equations I need help in finding all solutions in positive integers :
i) $\dfrac 1x+\dfrac 2y-\dfrac3z=1 $
ii) $\dfrac 1{x^2}+\dfrac 2{y^2}+\dfrac 3{z^2}=\dfrac 23$
iii) $p^x-y^3=1$ , where $p$ is a given prime
Please help
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For the 1st equation, we have $${1\over x}+{2\over y}>1$$ so we can't have both $x\ge3$ and $y\ge3$. If $x=2$ then $y$ must be 1, 2, or 3 for the inequality to hold; check these in turn, and find $(x,y,z)=(2,1,2),(2,2,6),(2,3,18)$. If $x=1$, you get $${2\over y}={3\over z}$$ which gives an infinity of solutions $(1,2k,3k)$. If $y=2$ then $${1\over x}={3\over z}$$ and you get $(k,2,3k)$. If $y=1$, then $3/z>1$ and all you get is $(2,1,2)$ again. So that should be it.
Similar methods apply to the second equation. We need $z\ge3$, $y\ge2$, $x\ge2$, but, aside from $(3,3,3)$, at least one of $x,y$ must be less than 3. Try all the cases --- there aren't that many.
For the third one, $$p^x=y^3+1=(y+1)(y^2-y+1)$$ so we must have $y+1=p^r$ and $y^2-y+1=p^s$ for some $r,s$. Then $$p^{2r}-3p^r=y^2-y-2=p^s-3$$ so $r=0$ or $s=0$ or $p=3$. It should be easy from there.
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i) We'll check two cases:
Case 1: $x\ge y$
Then we have:
$$1 = \frac 1x + \frac 2y - \frac 3z \le \frac 3y - \frac 3z$$ $$\frac 13 \le \frac{z-y}{yz}$$ $$3z - 3y \ge yz$$
Obviously $y\le2$, otherwise the inequality wouldn't hold.
If $\fbox{$y=1$}$ we get:
$$1 = \frac 1x + 2 - \frac 3z$$ $$\frac 3z - \frac 1x = 1$$ $$3x - z = zx$$
Obviosuly $z\le2$. plugging all possible values we get $\fbox{$x=z=2$}$.
If $\fbox{$y=2$}$ we get:
$$1 = \frac 1x + 1 - \frac 3z$$ $$\frac 3z = \frac 1x$$
Hence all solutions are: $\fbox{$(k,2,3k); k \in \mathbb{N}$}$
Simular method for $$y\le x$$
ii) Use Garry's hint and note that at at least one of the variables have to be divisible by 3.
iii) The equation is equivalent to:
$$p^x = (y+1)(y^2 - y + 1)$$
Now we have two cases:
**Case 1:**$(y+1, y^2 - y + 1) = 1$
This means that this numbers are coprime and that means that one of them has to be equal to one. Checking both cases yields: $$(p,x,y) = (2,1,1)$$
Case 2:$p$ divides both numbers
Then we have:
$$p\mid y^2 - y + 1 -(y+1) = y(y-2)$$
Since $p\mid y+1$ we get $p\not\mid y$. Hence $p\mid y-2$. Then we have: $p \mid y+1 - y + 2=3 \implies p = 3$
$y=1$ doesn't yield solution and $y=2$ gives us $p=3, x=2$. Now let $y\ge 3$. Obviously $y^2 - y + 1 > y + 1$ for $y>2$ and hence we have:
$$y+1 = 3^n \quad y^2 - y + 1 = 3^m, \quad m>n$$
Now check all possible combinations for y modulo 3 and you'll get a contradiction fairly easy.
i) Multiplying with xyz gives $$yz + 2xz -3xy = xyz$$
Beside (2,1,2) and (2,3,18) I only found solutions with x = 1 or y = 2, but I have no proof that there are no more. It is probably difficult to find all solutions here.
ii) Multiplying with $3x^2y^2z^2$ gives $$3y^2z^2 + 6x^2z^2 + 9x^2y^2 = 2x^2y^2z^2$$
I checked all triples with x,y,z$ \le 1000$ and (3,3,3) is the only solution. But I have no proof that there are no more solutions.
iii) Catalan's conjecture (now proven) states that the only nontrivial solution of $a^b-c^d=1$ is $3^2-2^3=1$ (nontrivial means a,b,c,d > 1)