0

I'm trying to solve the following equation in $\mathbb{Z^2}$ as i asked to do : $$(x+1)^2=9+5y$$ but actually this equation has more than two solutions ... what does $\mathbb{Z^2}$ stands for ?

Hedwig
  • 153

3 Answers3

1

Solving this equation is equivalent to find perfect squares congruent with $9$ (or better, with $4$) modulo $5$, that is $$z^2\equiv 4\pmod 5$$

This modular equation has two solutions in $\Bbb Z_5$, namely $z\equiv2$ and $z\equiv3$.

So the set of solutions of the equation is $$\left\{(x,y):x=5k+r, y=\frac{(x+1)^2-9}{5},k\in\Bbb Z,r\in\{1,2\}\right\}$$

ajotatxe
  • 65,084
  • as we can see here $x$ has two solutions and $y$ could have too, right ? – Hedwig May 25 '14 at 21:46
  • There are two solutions, that is, two values for $x$ and another two for $y$ for each value of the dummy variable $k$ in the set. Hence, there are infinitely many solutions. – ajotatxe May 26 '14 at 11:46
0

Hint: $(x+1)^2-3^2 = 5y$ or $(x+4)(x-2) = 5y$.

0

$\mathbb{Z^2}$ denotes the set of all ordered pairs of integers, e.g. (0,0), (4,2), (-1,3), and so on.

The equation does have an infinite number of solutions. The usual way to represent an answer in this case is to give $x$ and $y$ both in terms of some number $n$ (integer of course in this case).