Wikipedia:Reference desk/Archives/Mathematics/2025 March 3

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March 3

How to find a solution to this equation so the result is a perfect square without factorizing the semiprime ?

Simple question, I’ve the following expression : (y² + x×2032123)÷(17010411399424)

for example, x=2151695167965 and y=9 leads to 257049 which is the perfect square of 507

I want to find 1 or more set of integer positive x and y such as the end result is a perfect square (I mean where the square root is an integer). But how to do it if the divisor 17010411399424 is a different integer which thar time is non square and/or 2032123 is replaced by a semiprime impossible to factor ? 2A0D:E487:133F:E9BF:C9D5:9381:E57D:FCE8 (talk) 21:35, 3 March 2025 (UTC)

We can generalize to finding solutions to $(y^{2}+Ax)/B=z^{2}$ for fixed $A,B$ (in your case, 2032123 and 17010411399424 respectively.) Rearranging yields $y^{2}-Bz^{2}=Ax$. As long as there is some $y,z$ such that $y^{2}\equiv Bz^{2}\!\!{\pmod {A}}$, you can generate infinitely many solutions by taking $y'=y+mA$ and $z'=z+nA$ and working backwards to get $x$. Of course, some solutions correspond to negative values, but you can always just increase $y'$ and/or decrease $z'$ as needed. To find if there is such $y,z$ satisfying $y^{2}\equiv Bz^{2}\!\!{\pmod {A}}$ in the first place, you could just check values $y,z$ between 1 and $A$ inclusive until you find one, without needing to factorize. GalacticShoe (talk) 04:00, 4 March 2025 (UTC)

I need only positive solutions and where y<A

May you give a step by step example please?

Also, what do you mean by checkin values $y,z$ between 1 and $A$ inclusive until you find one ? How to do it ? Becuase I suppose that if A is 2000 bits long that this can t be done at random

2A0D:E487:35F:E1E1:51B:885:226F:140F (talk) 05:07, 4 March 2025 (UTC)