Testwiki:Reference desk/Archives/Mathematics/2016 November 15

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November 15

Solving diophantine equations

From my previous question, I read about diophantine equations. Now, I am looking at squares. As an example, I have x^2+38x+49. I want to find integer solutions for x that the equation produces a square. I know that x=0 produces 49, which is a square. I know that x=10 produces 23^2. I did that by hand. Is there an algorithm other than trying 0, 1, 2,... — Preceding unsigned comment added by 68.115.219.130 (talk) 15:42, 15 November 2016 (UTC)

If you make the substitution $x=t-19$ in your equation then you end up with the equation $t^2-312=y^2$. Rearranging gives $(t-y)(t+y)=312$. The left side is a factorization of 312 into two factors of the same parity; you can find all solutions by finding all factorizations of 312.--JBL (talk) 15:57, 15 November 2016 (UTC)

The factors can't be odd so it really comes down to factorizations of 78. It gets more interesting if the leading coefficient isn't 1. See Square triangular number for a notable example. --RDBury (talk) 20:29, 15 November 2016 (UTC)

See here. Count Iblis (talk) 01:34, 16 November 2016 (UTC)

That makes sense. I started with 3x13 and the solution is factoring 3x13x2x2x2. The article on square triangle numbers is very helpful. That is what my tutor asked. He told me some numbers can be arranged as both a triangle and a square. I don't know about the video. I listened, but I am blind, so it didn't make a lot of sense. — Preceding unsigned comment added by 68.115.219.130 (talk) 13:44, 16 November 2016 (UTC)