Diophantine Inequality 4 a x − y 2 + z &lt; 0For fixed positive integer a and positive z in 4 a x − y 2 + z &lt; 0 ,find how many integer solutions (x,y) there are as a function of z ∈ Z + , subject to constraints x ≥ 0 and a ≥ y ≥ 0.This form comes from the quadratic polynomial discriminant. I am not sure where to begin as this isn't the usual quadratic Diophantine inequality (or equation for that matter) that I find through search engines.

Question

Diophantine Inequality   4  a  x  −      y    2    +  z  &amp;lt;  0For fixed positive integer a and positive z in  4  a  x  −      y    2    +  z  &amp;lt;  0  ,find how many integer solutions (x,y) there are as a function of   z  ∈            Z        +  , subject to constraints   x  ≥  0 and   a  ≥  y  ≥  0.This form comes from the quadratic polynomial discriminant. I am not sure where to begin as this isn&#039;t the usual quadratic Diophantine inequality (or equation for that matter) that I find through search engines.

Raven Hawkins · Accepted Answer

Step 1We can rewrite the starting inequality as   x  &amp;lt;                    y        2            −      z              4      a      . Since we also know that   x  ≥  0, this can be rewritten to  0  ≤  x  &amp;lt;                    y        2            −      z              4      a        ,  0  ≤  y  ≤  aNote that y can be further restricted to       z    &amp;lt;  y  ≤  a. This is because if   y  ≤      z  , then the upper bound on x would not be positive.Step 2From here, we can go through the possible y and add:      ∑          y      =      ⌊              z            ⌋      +      1        a        ⌈                            y          2                −        z                    4        a              ⌉  This is the final answer for the number of lattice points in the region.

Diophantine Inequality 4ax-y^2+z<0

Answered question

Answer & Explanation

New Questions in Algebra I