# Parametric solution of the Diophantine equation $\frac{1}{p}=\frac{1}{x}+\frac{1}{y}+\frac{1}{z} ,x,y,z∈Z^+.$

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 ?

## 2 Answers

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:~$ =-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-= 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:~$ =-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-= • I did not understand. You that have copied here a decision from there? math.stackexchange.com/questions/419766/… – individ Sep 11 '14 at 8:03 • @individ, I did not look at the links you gave. So, no copying. i – Will Jagy Sep 11 '14 at 16:51 • @WillJagy: Thank you for your nice Solution and what do you think about the Second Diophantine Equation? – Bumblebee Sep 12 '14 at 4:16 • @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. – Will Jagy Sep 12 '14 at 4:35 • @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? – Bumblebee Sep 12 '14 at 5:58 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\$

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