• #### ( x ^ 2 + y ^ 2 + z ^ 2 = xyz ) for positive integers (x, y, z)

From henhanna@gmail.com@21:1/5 to All on Sun Jun 26 18:13:51 2022
x ^ 2 + y ^ 2 + z ^ 2 = xyz

1. Show some positive integers (x, y, z) that satisfy this equation

2. How many positive-integer triples (x, y, z) satisfy this equation ?

pls wait a few days before posting answers or hints.

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• From Ilan Mayer@21:1/5 to henh...@gmail.com on Thu Jun 30 05:34:42 2022
On Sunday, June 26, 2022 at 9:13:53 PM UTC-4, henh...@gmail.com wrote:
x ^ 2 + y ^ 2 + z ^ 2 = xyz

1. Show some positive integers (x, y, z) that satisfy this equation

2. How many positive-integer triples (x, y, z) satisfy this equation ?

pls wait a few days before posting answers or hints.
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Some solutions:
3 3 3
3 3 6
3 6 15
3 15 39
3 39 102
3 102 267
3 267 699
6 15 87
6 87 507
15 39 582

There is an infinite number of solutions.
If a,b,c is a solution (a^2+b^2+c^2 = a*b*c), then a,c,a*c-b is also a solution Proof: a^2+c^2+(a*c-b)^2 = a^2+c^2+a^2*c^2+b^2-2*a*b*c = a^2*c^2-a*b*c = a*c*(a*c-b)

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__/\\ //\__ Ilan Mayer
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• From henhanna@gmail.com@21:1/5 to Ilan Mayer on Thu Jun 30 14:14:15 2022
On Thursday, June 30, 2022 at 5:34:45 AM UTC-7, Ilan Mayer wrote:
On Sunday, June 26, 2022 at 9:13:53 PM UTC-4, henh...@gmail.com wrote:
x ^ 2 + y ^ 2 + z ^ 2 = xyz

1. Show some positive integers (x, y, z) that satisfy this equation

2. How many positive-integer triples (x, y, z) satisfy this equation ?

pls wait a few days before posting answers or hints.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.

Some solutions:
3 3 3
3 3 6
3 6 15
3 15 39
3 39 102
3 102 267
3 267 699
6 15 87
6 87 507
15 39 582

There is an infinite number of solutions.
If a,b,c is a solution (a^2+b^2+c^2 = a*b*c), then a,c,a*c-b is also a solution
Proof: a^2+c^2+(a*c-b)^2 = a^2+c^2+a^2*c^2+b^2-2*a*b*c = a^2*c^2-a*b*c = a*c*(a*c-b)

all multiples of 3 ... but never 9 ? or 12 ?

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Ilan Mayer@21:1/5 to henh...@gmail.com on Fri Jul 1 06:40:08 2022
On Thursday, June 30, 2022 at 5:14:18 PM UTC-4, henh...@gmail.com wrote:
On Thursday, June 30, 2022 at 5:34:45 AM UTC-7, Ilan Mayer wrote:
On Sunday, June 26, 2022 at 9:13:53 PM UTC-4, henh...@gmail.com wrote:
x ^ 2 + y ^ 2 + z ^ 2 = xyz

1. Show some positive integers (x, y, z) that satisfy this equation

2. How many positive-integer triples (x, y, z) satisfy this equation ?

pls wait a few days before posting answers or hints.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.

Some solutions:
3 3 3
3 3 6
3 6 15
3 15 39
3 39 102
3 102 267
3 267 699
6 15 87
6 87 507
15 39 582

There is an infinite number of solutions.
If a,b,c is a solution (a^2+b^2+c^2 = a*b*c), then a,c,a*c-b is also a solution
Proof: a^2+c^2+(a*c-b)^2 = a^2+c^2+a^2*c^2+b^2-2*a*b*c = a^2*c^2-a*b*c = a*c*(a*c-b)

all multiples of 3 ... but never 9 ? or 12 ?

As mentioned above, if a,b,c is a solution (a^2+b^2+c^2 = a*b*c), then a,c,a*c-b is also a solution

In the above solutions all numbers are divisible by 3, but not by 9.
If a = 3*x, b = 3*y and c = 3*z, where x, y, and z are not divisible by 3, then a*c-b = 3*(3*x*z-y).
3*x*z-y in not divisible by 3 as well, and so a*c-b is divisible by 3 but not divisible by 9.
This means that in all solutions derived from the ones above the numbers are divisible by 3 but not by 9.

In the above solution odd numbers have the form 4*n+3 and even numbers have the form 4*n+2.
If a, b, c are all odd, then a*c has the form 4*n+1 ((4*x+3)*(4*y+3) = 4*(4*x*y+3*x+3*y+2)+1).
Then a*c-b has the form 4*n+2 ((4*x+1)-(4*y+3) = 4*(x-y-1)+2).
If b is even, then a*c-b has the form 4*n+3 ((4*x+1)-(4*y+2) = 4*(x-y-1)+3).
If a or c is even, then a*c has the form 4*n+2 ((4*x+2)*(4*y+3) = 4*(4*x*y+3*x+2*y+1)+2).
Then a*c-b has the form 4*n+3 ((4*x+2)-(4*y+3) = 4*(x-y-1)+3).
This means that in all solutions derived from the ones above even numbers have the form 4*n+2, and therefore are not divisible by 12.