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Is there a general solution to $n (a)^2=b^2+c^2+1$?

Where $n$ is an integer and $a,b,c$ are rational numbers?

I am looking for all rational solutions.

Arturo Magidin
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Darrin Taylor
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  • What kind of an answer are you expecting. This equation is relatively easy to fulfill. – Jyrki Lahtonen Jul 08 '13 at 10:29
  • in the case of x^2+y^2-z^2==0 the solutions are given by x=2 rs , y = r^2-s^2, z= r^2+s^2 this gives all unique factor free solutions. Alternately x=a+k, y=k^2/(2a)+k, z=a+k^2/(2a)+k. In this way all rational solutions can be generated. Is there a similar way to generate ALL solutions not just to say that a single solution can be found? – Darrin Taylor Jul 08 '13 at 16:14
  • @DarrinTaylor Perhaps you are not asking for ALL rational solutions, but a general solution for $x,y,z$ given any fixed choice of $n$? This isn't clear from the way the question is written which does not make any distinction between parameters and variables. – Erick Wong Feb 24 '14 at 15:44

2 Answers2

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Very little to this. The variables $b$ and $c$ can be anything you want them to be. The rational $a$ needs to be chosen in such a way that $$ \frac{b^2+c^2+1}{a^2} $$ is an integer. Basically we want the denominator of $a$ to cancel (when squared) denominator of $b^2+c^2+1$, and the numerator of $a$ is severely constrained. It will be restricted to $1$ in most cases.

Jyrki Lahtonen
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For some idea of what you're up against, see What is going wrong with my solution for Project Euler 224? which just deals with the case $n=1$, and just with integer values for $a,b,c$. In summary, OP finds an infinity of solutions by taking $a=2m^2+1$, $b=2m$, $c=2m^2$, but notes that there are lots of solutions not of this form. User Peter comments that for $1\le b\le c\le10^5$ there are more than $4329$ solutions in integers. User Peter also notes that there's another infinity of integer solutions given by $a=10x^2+4x+3$, $b=6x^2+4x+2$, $c=8x^2+2x+2$.

The question of finding integer solutions of $a^2=b^2+c^2+1$ is also studied in some detail at Integral solutions of $x^2+y^2+1=z^2$

Gerry Myerson
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