Mathematics desk
< September 24	<< Aug \| September \| Oct >>	September 26 >

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is a transcluded archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

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)[reply]

If you want integer solutions {x,y} then these only exist if conditions are met for the constants, see Diophantine_equation#One_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 [1]. 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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]