\part{Elementary approach to Diophantine equations}



<h3 class=dred>Diophantine equations</h3>
<p>Equations with multiple unknown variables that have solutions in \(\mathbb{Z}\)</p>

<h3 class=dred>Linear Diophantine Equation (LDE)</h3>
<p>Diophantine equations with linear variables</p>
<p>\( : a,b,c \in \mathbb{Z}\)</p>
<p>\(ax+by=c : a,b,c \in \mathbb{Z}\)</p>
<h3 class=dred>Linear Diophantine Equation (LDE)</h3>
<p>\( a=0  \land b|c \implies y= \frac{c}{b} \)</p>
<p>\( \neg (\gcd(a,b) | c) \implies \text{ no solutions } \)</p>
<ul>
	<li>Compute \( \gcd(a,b) \) </li>
	<li>\(x = x_0 + \frac{b}{\gcd(a,b)}n \)</li>
	<li>\(y = y_0 - \frac{a}{\gcd (a,b)}n \)</li>
</ul>
<p>Such that \( (x_0,y_0) \) are the results found from extended euclidean algorithm </p>

<h3 class=dred>Second degree Diophantine equations</h3>
<p>Diophantine equations including degree 2 polynomials</p>
<p>\(x^2+y^2=a : a \in \mathbb{Z}\)</p>
<p>\(\nexists a : x^2 + y^2 = a \iff a \equiv 3 \mod 4 \)</p>
<p>All even numbers squared reduce to 0 modulo 4 since \( (2n)^2 = 4n^2 \equiv 0 \mod 4 \)  and odd numbers 1 modulo 4 since \( (2n-1)^2 = 4n^2 -4n +1 = 4(n^2 -n) +1 \equiv 1 \mod 4  \). Therefore \( x^2 + y^2 \) can never reduce to 3 modulo 4 since the possible combinations are \(0+0=0,0+1=1,1+0=1,1+1=2\), excluding 3 from the possibilities. Therefore \(a \equiv 3 \mod 4\) shows a disparity between the capabilities of the sum of two squares and \(a\), proving the unsatisfiability of this diophantine equation.</p>
<h3 class=dred>Pell's equation</h3>
<p>\(x^2 -ny^2 = 1\)</p>




\begin{theorem}[Euclid's formula]
\[a=m^2 + n^2 \]
\[b=2nm \]
\[c=m^2 - n^2 \]
\end{theorem}
