Let y^2=xz-x^2+1, and if value of x is known, can y and z be directly calculated? Given all variable

[MARTHA]

Let y^2=xz-x^2+1, and if value of x is known, can y and z be directly calculated? Given all variables are Integers.

[science_man_88]

Originally Posted by MARTHA

Let y^2=xz-x^2+1, and if value of x is known, can y and z be directly calculated? Given all variables are Integers.

Is this homework ?

[LaurV]

Originally Posted by MARTHA

Let y^2=xz-x^2+1, and if value of x is known, can y and z be directly calculated? Given all variables are Integers.

Yes. There are an infinite amount of them. You can write your expression like
xz=y^2+x^2-1, or (divide both left and right by x) \(z=\frac{y^2+x^2-1}x\), or \(z=\frac{y^2-1}x+x\). Now you see that the only condition to have z integer is that y^2-1 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^2-1 will end in 0 or 5.

[a1call]

Wolfram Alpha link:

https://www.wolframalpha.com/input/?...%2B+1+for+y,+z

[LaurV]

whaaa, haha

[a1call]

Originally Posted by LaurV

whaaa, haha
Attachment 17790

It's not what it seems. It's not a case of Pet-Discrimination.
Dog-Health is more Number-Theory related.

[LaurV]

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)

[S485122]

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

[a1call]

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

[Thomas11]

Following LaurV's suggestion the following solution will be found:

\(y = k\cdot x + 1\)
\(z = 2k + (k^2+1)\cdot x\)

for all \(k\in N\)

[MARTHA]

Thank you all !!!
you guys are amazing