Difference of Square Roots

[MARTHA]

Dear All,

I am looking for the solution of equation x^0.5 - y^0.5 = z; where x and y are integers while z is any floating number.. If z is known, can x and y be directly calculated rather then brute force?

[R. Gerbicz]

Because z is a floating point number this is a perfect solution, if you want a solution of absolute precision 2^(-b).

(10:34) gp > ff(z,b)={y=2^(2*b);x=round((2^b+z)^2);return([x,y])}
%1 = (z,b)->y=2^(2*b);x=round((2^b+z)^2);return([x,y])
(10:34) gp > ff(600.8,52)
%2 = [20282409603657081949259500034970, 20282409603651670423947251286016]
(10:34) gp > x=20282409603657081949259500034970;
(10:34) gp > y=20282409603651670423947251286016;
(10:34) gp > d=sqrt(x)-sqrt(y)
%5 = 600.79999999999999997335463576229321828
(10:35) gp > (d-600.8)*2^52
%6 = -0.12000005245208740234
(10:35) gp >

ps. using symmetry you can assume that z>=0 (just swap x and y).

[MARTHA]

Originally Posted by R. Gerbicz

Because z is a floating point number this is a perfect solution, if you want a solution of absolute precision 2^(-b).

(10:34) gp > ff(z,b)={y=2^(2*b);x=round((2^b+z)^2);return([x,y])}
%1 = (z,b)->y=2^(2*b);x=round((2^b+z)^2);return([x,y])
(10:34) gp > ff(600.8,52)
%2 = [20282409603657081949259500034970, 20282409603651670423947251286016]
(10:34) gp > x=20282409603657081949259500034970;
(10:34) gp > y=20282409603651670423947251286016;
(10:34) gp > d=sqrt(x)-sqrt(y)
%5 = 600.79999999999999997335463576229321828
(10:35) gp > (d-600.8)*2^52
%6 = -0.12000005245208740234
(10:35) gp >

ps. using symmetry you can assume that z>=0 (just swap x and y).

Thanks for your quick response, but being a layman I am not able to understand your code..
Could you please explain in simple english the solution of this equation for example:
x^0.5-y^0.5=1.234567

Thanks again

[axn]

Originally Posted by MARTHA

I am looking for the solution of equation x^0.5 - y^0.5 = z; where x and y are integers while z is any floating number.. If z is known, can x and y be directly calculated rather then brute force?

Isn't this trying to map a countably infinite set to an uncountably infinite one?

[firejuggler]

hmmm since X and Y are integer and different from 0, wouldn't that seriously reduce the number of solution?

[Nick]

Originally Posted by MARTHA

Could you please explain in simple english the solution of this equation for example:
x^0.5-y^0.5=1.234567

Thanks again

There is no solution in integers to that equation.
If z is a non-zero rational number then \(\sqrt{x}\) and \(\sqrt{y}\) must be integers.

[MARTHA]

Originally Posted by Nick

There is no solution in integers to that equation.
If z is a non-zero rational number then \(\sqrt{x}\) and \(\sqrt{y}\) must be integers.

Thanks to the formula given by Sir Gerbicz, we have
\(\sqrt{281475018135852}\) - \(\sqrt{281474976710656}\)=1.234567001
But there must be lower values of x and y for this..

For example if z=1.026, we have \(\sqrt{1050678}\)- \(\sqrt{1048576}\)=1.02585333 with formula
but \(\sqrt{24}\)- \(\sqrt{15}\)=1.02599 would be a better approximation...

[LaurV]

You asked for solution not for approximation.
One way to see is if you amplify with the conjugate. \((\sqrt x -\sqrt y)(\sqrt x +\sqrt y)=z(\sqrt x +\sqrt y)\) then you have \(x-y=z(\sqrt x +\sqrt y)\). The left side is an integer, the right side is not, except very particular cases, like for example, z is irrational, and it has a common square free part with x and y, see \(\sqrt{75}-\sqrt{12}=3*\sqrt 3\). For z non-integer, rational, there is no solution with square free parts, as Nick said.

[Dr Sardonicus]

Originally Posted by LaurV

You asked for solution not for approximation.
One way to see is if you amplify with the conjugate. \((\sqrt x -\sqrt y)(\sqrt x +\sqrt y)=z(\sqrt x +\sqrt y)\) then you have \(x-y=z(\sqrt x +\sqrt y)\). The left side is an integer, the right side is not, except very particular cases, like for example, z is irrational, and it has a common square free part with x and y, see \(\sqrt{75}-\sqrt{12}=3*\sqrt 3\). For z non-integer, rational, there is no solution with square free parts, as Nick said.

To be fair, the OP did specify z as a "floating number," which I interpret as a "floating point number," which is not an exact numeric type -- it has limited precision. R. Gerbicz's routine incorporates the precision into the answer.

Trying to find small solutions -- that's been a real head-scratcher for me. I note that, if z is a positive integer k, x = (n + k)^2 and y = n^2 give exact integer solutions. If z is not an integer, the integer part of z pretty much pegs the difference between the integer parts of sqrt(x) and sqrt(y).

[MARTHA]

Thank you all for your precious time...
Its quite fascinating that even with Computers, there is still so much to be discovered in the field of MATH..

[science_man_88]

Originally Posted by MARTHA

Thank you all for your precious time...
Its quite fascinating that even with Computers, there is still so much to be discovered in the field of MATH..

https://en.m.wikipedia.org/wiki/List...ematics_topics here's a list of what is covered ...