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I have prove that, for any given positive integer $p,$ parametric solution of the Diophantine equation $$\frac{1}{p}=\frac{1}{x}+\frac{1}{y}$$ can be written in the form $x=ac(a+b),y=bc(a+b),$ where $p=abc.$
Proof
Let $\frac{1}{p}=\frac{1}{x}+\frac{1}{y} ,x,y∈Z^+.$
Then $x+y=t$ and $xy=pt$ for some $t∈Z^+.$
Now the quadratic equation $z^2-tz+pt=0$ has two integer roots $x,y.$
Discriminant of this equation can be written as $Δ_(x,y)=t^2-4pt=q^2, q∈Z^+.$
The quadratic equation $t^2-4pt-q^2=0$ gives the value of $t.$
$Δ_t=16p^2+4q^2=4r^2,r∈Z^+.$
$4p^2+q^2=r^2,r∈Z^+.$
This equation is of the form of Pythagoras equation.
Therefore $p=abc,q=(a^2-b^2 )c$ and $r=(a^2+b^2 )c$ where $a,b,c$ are parameters.
Backward substitution gives that $t=(a+b)^2 c.$
Hence we can obtain that
$x=ac(a+b),y=bc(a+b).$

Then I was try to find the general parametric solution of the Diophantine equation $$\frac{1}{p}=\frac{1}{x}+\frac{1}{y}+\frac{1}{z} ,x,y,z∈Z^+.$$ I have found some particular solutions like, $$\frac{1}{n}=\frac{1}{n+2}+\frac{1}{n(n+1)}+\frac{1}{(n+1)(n+2)}$$ $$\frac{1}{n}=\frac{1}{n+1}+\frac{1}{n(n+2)}+\frac{1}{n(n+1)(n+2)}$$ $$\frac{1}{n}=\frac{1}{n+1}+\frac{1}{(n+1)^2} +\frac{1}{n(n+1)^2 }$$ $$\frac{1}{n}=\frac{1}{n+1}+\frac{1}{n(2n+1)}+\frac{1}{(n+1)(2n+1)}$$ $$\frac{1}{n}=\frac{1}{n+1}+\frac{1}{(n^2+n+1)}+\frac{1}{n(n+1)(n^2+n+1)}.$$ But still I have no idea about how to attack the general one.
Here I have three questions.
1) Is there any different proof for general solution of first equation than my proof ?
2) Is there any general parametric solution for the second Diophantine equation ?
3) Is there any reference for these type of Diophantine equations ?

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Got it. Your equation is $$ xy = px + py, $$ $$ xy - px - py = 0, $$ $$ xy - px - py + p^2 = p^2, $$ $$ (x-p)(y-p) = p^2. $$ Apparently this observation occurs at Number of solution for $xy +yz + zx = N$

All solutions are given by finding a divisor $w$ of $p^2,$ with triple $$ \color{magenta}{ \left( p, \; \; p + w, \; \; p + \frac{p^2}{w} \; \right).} $$ If $w < p$ these are in order, if $w=p$ it is just $(p,2p,2p),$ if $w > p$ it is a repeat but out of order. So, the total number of solutions is $$ \frac{1 + d(p^2)}{2}, $$ where $d(n)$ is the number of positive divisors of $n.$

Note that the primitive triples, $\gcd(p,x,y),$ come when my $w$ is $1$ or some other square, so $p^2/w$ is also a square, in addition we require $\gcd(w,p^2/w)= 1$; for example $(6,10,15)$ with $w=4$ and $p^2/w = 9.$

OR $$ (30,31,930); \; \; (30,34,255); \; \; (30,39,130); \; \; (30,55,66). $$

$p$ up to $30.$

      p      x      y
      1      2      2
      2      4      4
      2      3      6
      3      6      6
      3      4     12
      4      8      8
      4      6     12
      4      5     20
      5     10     10
      5      6     30
      6     12     12
      6     10     15
      6      9     18
      6      8     24
      6      7     42
      7     14     14
      7      8     56
      8     16     16
      8     12     24
      8     10     40
      8      9     72
      9     18     18
      9     12     36
      9     10     90
     10     20     20
     10     15     30
     10     14     35
     10     12     60
     10     11    110
     11     22     22
     11     12    132
     12     24     24
     12     21     28
     12     20     30
     12     18     36
     12     16     48
     12     15     60
     12     14     84
     12     13    156
     13     26     26
     13     14    182
     14     28     28
     14     21     42
     14     18     63
     14     16    112
     14     15    210
     15     30     30
     15     24     40
     15     20     60
     15     18     90
     15     16    240
     16     32     32
     16     24     48
     16     20     80
     16     18    144
     16     17    272
     17     34     34
     17     18    306
     18     36     36
     18     30     45
     18     27     54
     18     24     72
     18     22     99
     18     21    126
     18     20    180
     18     19    342
     19     38     38
     19     20    380
     20     40     40
     20     36     45
     20     30     60
     20     28     70
     20     25    100
     20     24    120
     20     22    220
     20     21    420
     21     42     42
     21     30     70
     21     28     84
     21     24    168
     21     22    462
     22     44     44
     22     33     66
     22     26    143
     22     24    264
     22     23    506
     23     46     46
     23     24    552
     24     48     48
     24     42     56
     24     40     60
     24     36     72
     24     33     88
     24     32     96
     24     30    120
     24     28    168
     24     27    216
     24     26    312
     24     25    600
     25     50     50
     25     30    150
     25     26    650
     26     52     52
     26     39     78
     26     30    195
     26     28    364
     26     27    702
     27     54     54
     27     36    108
     27     30    270
     27     28    756
     28     56     56
     28     44     77
     28     42     84
     28     36    126
     28     35    140
     28     32    224
     28     30    420
     28     29    812
     29     58     58
     29     30    870
     30     60     60
     30     55     66
     30     50     75
     30     48     80
     30     45     90
     30     42    105
     30     40    120
     30     39    130
     30     36    180
     30     35    210
     30     34    255
     30     33    330
     30     32    480
     30     31    930
jagy@phobeusjunior:~$

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All primitive solutions are given by finding a divisor $w$ of $p$ such that $\gcd(w,p/w) = 1$ with triple $$ \color{magenta}{ \left( p, \; \; p + w^2, \; \; p + \frac{p^2}{w^2} \; \right).} $$ To keep them ordered we also choose $w \leq \sqrt p.$ If $p$ is a square in the first place, larger than $1,$ then $w=\sqrt p$ does not ever give a primitive solution anyway, that just gives $(p,2p,2p).$

Here are just the primitive ones for $p \leq 30$ and then $p=210.$

      p      x      y
      1      2      2
      2      3      6
      3      4     12
      4      5     20
      5      6     30
      6      7     42
      6     10     15
      7      8     56
      8      9     72
      9     10     90
     10     11    110
     10     14     35
     11     12    132
     12     13    156
     12     21     28
     13     14    182
     14     15    210
     14     18     63
     15     16    240
     15     24     40
     16     17    272
     17     18    306
     18     19    342
     18     22     99
     19     20    380
     20     21    420
     20     36     45
     21     22    462
     21     30     70
     22     23    506
     22     26    143
     23     24    552
     24     25    600
     24     33     88
     25     26    650
     26     27    702
     26     30    195
     27     28    756
     28     29    812
     28     44     77
     29     30    870
     30     31    930
     30     34    255
     30     39    130
     30     55     66
jagy@phobeusjunior:~$

    210    211  44310
    210    214  11235
    210    219   5110
    210    235   1974
    210    246   1435
    210    259   1110
    210    310    651
    210    406    435
jagy@phobeusjunior:~$ 

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  • $\begingroup$ I did not understand. You that have copied here a decision from there? math.stackexchange.com/questions/419766/… $\endgroup$ – individ Sep 11 '14 at 8:03
  • $\begingroup$ @individ, I did not look at the links you gave. So, no copying. i $\endgroup$ – Will Jagy Sep 11 '14 at 16:51
  • $\begingroup$ @WillJagy: Thank you for your nice Solution and what do you think about the Second Diophantine Equation? $\endgroup$ – Bumblebee Sep 12 '14 at 4:16
  • $\begingroup$ @Nilan, it seems unlikely that the version with three fractions on the right hand side can be handled adequately by parametrizations, otherwise I think something would have been included at en.wikipedia.org/wiki/Erd%C5%91s%E2%80%93Straus_conjecture Meanwhile, the prcise trick that i used here comes out to $$ (x-p)(y-p)(z-p) = p^2 (x+y+z - p), $$ which probably allows speedy computer search but does not show me a way to find all solutions. I suggest you do a computer search for all primitive soutions with $p$ up to some reasonable bound. $\endgroup$ – Will Jagy Sep 12 '14 at 4:35
  • $\begingroup$ @WillJagy, Thanks. I'm also interested about the equation $\frac{1}{p}=\frac{1}{x^2}+\frac{1}{y^2}$ and higher powers. I know for some $p$ there are no solutions. Do you know some thin about this? $\endgroup$ – Bumblebee Sep 12 '14 at 5:58
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For the equation: $$\frac{1}{p}=\frac{1}{x^2}+\frac{1}{y^2}$$

We will use the formula in the link record of decision in another form:

$$p=\sqrt{ks}$$

$$x^2=\sqrt{k}(\sqrt{k}+\sqrt{s})$$

$$y^2=\sqrt{s}(\sqrt{s}+\sqrt{k})$$

It is clear that:

$$k=a^4$$

$$s=b^4$$

And this is the Pythagorean triples. $a^2+b^2=c^2$

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  • $\begingroup$ oh....... Wonderfull. Thank you very much. You are a master of Diophantine Equations :) $\endgroup$ – Bumblebee Sep 12 '14 at 8:21

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