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20180302, 03:31  #1 
Jan 2018
43 Posts 
Let y^2=xzx^2+1, and if value of x is known, can y and z be directly calculated? Given all variable
Let y^2=xzx^2+1, and if value of x is known, can y and z be directly calculated? Given all variables are Integers.

20180302, 03:36  #2 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 

20180302, 05:12  #3  
Romulan Interpreter
"name field"
Jun 2011
Thailand
2^{4}·613 Posts 
Quote:
xz=y^2+x^21, or (divide both left and right by x) \(z=\frac{y^2+x^21}x\), or \(z=\frac{y^21}x+x\). Now you see that the only condition to have z integer is that y^21 is divisible by x. Pick any x, say x=5, then all y ending in 1, 4, 6, or 9 will have squares ending in 1 or 6, so y^21 will end in 0 or 5. 

20180302, 05:21  #4 
"Rashid Naimi"
Oct 2015
Remote to Here/There
2·1,087 Posts 

20180302, 05:35  #5 
Romulan Interpreter
"name field"
Jun 2011
Thailand
2^{4}·613 Posts 

20180302, 05:53  #6  
"Rashid Naimi"
Oct 2015
Remote to Here/There
2×1,087 Posts 
Quote:
DogHealth is more NumberTheory related. 

20180302, 06:31  #7 
Romulan Interpreter
"name field"
Jun 2011
Thailand
9808_{10} Posts 
I am innocent I just followed your link. However, for the top of my head I can not imagine what kind of association wolfram alfa did (I never searched for cats and dogs on my computers, either)

20180302, 06:40  #8 
"Jacob"
Sep 2006
Brussels, Belgium
2^{2}×439 Posts 
There is not enough data to DIRECTLY calculate y and z when x is known, even given the fact that all are integers, because there is not a unique solution.
Jacob Last fiddled with by S485122 on 20180302 at 06:47 Reason: removed the part doing the homework 
20180302, 07:00  #9 
"Rashid Naimi"
Oct 2015
Remote to Here/There
2·1,087 Posts 
Forgot to put the integer constraint.
Solution doesn't really change but for whatever reason the engine also prints the "Implicit derivatives" Whatever that is. https://www.wolframalpha.com/input/?...r+the+integers 
20180302, 09:59  #10 
Feb 2003
1915_{10} Posts 
Following LaurV's suggestion the following solution will be found:
\(y = k\cdot x + 1\) \(z = 2k + (k^2+1)\cdot x\) for all \(k\in N\) 
20180302, 12:37  #11 
Jan 2018
53_{8} Posts 
Thanks
Thank you all !!!
you guys are amazing 
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