$ax+by=c$
$x,y\in\mathbb{Z}$
Number Theory Hub
Formal Science // Polynomials

DIOPHANTINE
EQUATIONS

Polynomial equations where we demand a strict, uncompromising condition: only integer solutions are allowed. It is the mathematical equivalent of navigating a continuous world using only discrete, snapping grid points.

The Discrete Grid

Standard algebra allows for continuous solutions. If you graph the line $2x+4y=7$, there are an infinite number of points on that line. However, if we view this as a Linear Diophantine Equation, we are asking a different question: does this line ever cross an intersection of exact, whole numbers?

Bézout's Identity proves that the equation $ax+by=c$ has an integer solution if, and only if, $c$ is a multiple of the Greatest Common Divisor of $a$ and $b$.

Because $\gcd(2,4)=2$, and $7$ is not divisible by $2$, the line $2x+4y=7$ will slip through the integer grid forever, never once touching a perfect coordinate.

Fermat's Last Theorem

While linear Diophantine equations are relatively easy to solve using the Extended Euclidean Algorithm, non-linear ones can be historically difficult.

We know there are infinite integer solutions to $x^2+y^2=z^2$ (these are Pythagorean Triples, like $3^2+4^2=5^2$). But in 1637, Pierre de Fermat scribbled in the margin of a book that no three positive integers can satisfy the equation $a^n+b^n=c^n$ for any integer value of $n>2$.

"I have a truly marvelous demonstration of this proposition which this margin is too narrow to contain." — Pierre de Fermat

It took mathematicians over 350 years to finally prove he was right.

Linear Diophantine Solver

$ax + by = c$

3x+6y=12
Coefficient $a$3
Coefficient $b$6
Target $c$12
System Diagnostic
$\gcd(3, 6) = 1$
No Integer SolutionBecause 12 is NOT a multiple of 1, this line slips entirely through the integer grid without hitting a single perfect coordinate.