Wikipedia:Reference desk/Archives/Mathematics/2014 September 25

= September 25 =

Linear equation perfect square solutions
I was reading about finding a solution to a quadratic equations which is a perfect square (the solution is easy to understand) and I started wondering about linear equations. Given a simple linear equation in the form y=ax+b where a and b are constants, what value of x will ensure that y is a perfect square? I've tried turning ax+b into (cx+d)(cx+d), but that hasn't worked. — Preceding unsigned comment added by 209.149.115.99 (talk) 18:19, 25 September 2014 (UTC)


 * If you want integer solutions {x,y} then these only exist if conditions are met for the constants, see Diophantine_equation. Then you would just have to search among the (countable) solution set for perfect squares. You could also write y=z^2 and think of z^2=ax+b as a quadratic Diophantine equation, and solve using a Pell equation as described here . Both of these methods would miss solutions where x is not an integer and y is a perfect square. SemanticMantis (talk) 19:12, 25 September 2014 (UTC)


 * I worked on this last night in earnest, but I don't see the relationship between y2=ax+b and the Diophantine ax2+by2=k. I changed my original request of y to y2 to reinforce that I am looking for a square. However, the x is not a square. If I rearrange it, I can make, at best, ax+by2=k. Without the squared x, I cannot make any headway on the solutions provided. 209.149.115.99 (talk) 11:26, 26 September 2014 (UTC)
 * Sorry about that, I misread my mathworld link. I don't know much about this area, but you could try looking up the Ito (1987) paper cited at mathworld, and also see what other papers have cited it. SemanticMantis (talk) 14:51, 26 September 2014 (UTC)


 * If you're happy with just rational solutions then draw a circle around the origin intersecting where y=ax+b is zero. This will be a rational point since y is zero and a and b are rational so the other point on the circle where y=ax+b cuts it is also rational. Any choice of a rational number for x and 0 for y for the origin will do the same thing. Dmcq (talk) 12:03, 26 September 2014 (UTC) Oops silly me I misread the question, I'll have a think about it. Dmcq (talk) 17:08, 26 September 2014 (UTC)
 * If you are looking for integer solutions with a and b being integers then your question is equivalent to asking if $$y^2=b \mod a$$ . See the article Quadratic reciprocity for how the Legendre symbol can be used to quickly check if there is a solution is a or b are large - though it won't give an actual solution unfortunately. The article quadratic residue gives a ways of calculating the residues without checking each possibility except when a prime factor is of the form $$a=8n+1$$. Dmcq (talk) 17:29, 26 September 2014 (UTC)


 * Thanks. I didn't look at it as a mod problem. Once you throw that in, I can see why there is no simple solution for x. It looks like you'd have to try every x in the range you like to see if any work. 209.149.115.99 (talk) 17:49, 26 September 2014 (UTC)