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Use the remainder form of the Euclidean algorithm to compute the greatest common divisor, then divide both components of the fraction by it:The loop invariant is $a>0$, $b\ge 0$, and $\gcd (a,b)=\gcd (den,num)$. Initially it holds by symmetry of the greatest common divisor. For $b>0$, the integer division identity $a=qb+r$, with $0\le r<b$, shows that $a,b$ and $b,r$ have exactly the same common integer divisors. Thus the update preserves the loop invariant. The loop variant is the nonnegative integer $b$, which strictly decreases on each iteration. At exit $b=0$, so $a=\gcd (den,num)>0$. Dividing by $a$ preserves the rational number and makes the returned components coprime. In particular, zero is returned as $(0,1)$. No multiplication is needed, and all remainders remain within the original nonnegative Int range.

The Euclidean algorithm givesso $\gcd (754,91)=\gcd (26,91)=13$. Since $13\mid 39$, the criterion for a linear congruence guarantees a solution. One can also exhibit it directly: $26\cdot 5+91(-1)=39$, or in the original coefficients,Thus $x=5$ is a solution, with an explicit Bézout coefficient certificate.

The Euclidean algorithm givesThe last nonzero remainder is $g=92$. Back-substitution produces its Bézout coefficients:Thus $(m,n)=(-8,11)$ is a requested pair. The final equality independently checks the calculation.

The Euclidean algorithm repeatedly replaces $(r_{i-1},r_i)$ by $(r_i,r_{i-1}-q_i{r}_i)$, where the new remainder lies between $0$ and $r_i-1$. Common divisors are unchanged by this subtraction, and decreasing positive remainders ensure termination. The last nonzero remainder is the greatest common divisor.

The extended Euclidean algorithm also tracks coefficients $r_i=u_{i}a+v_{i}b$, beginning with coefficient pairs $(1,0)$ and $(0,1)$ and applying the same subtraction to the pairs. This invariant produces Bézout coefficients for the last nonzero remainder. Here the divisions areBack substitution givesThus $\gcd (580,252)=4$, with $m=10$ and $n=-23$ in Bézout's identity.

For an integer $n$ to be factored, John Pollard's method chooses an integer $a$ coprime to $n$ and an exponent $M$ divisible by many small prime powers. Compute $d=\gcd (a^M-1,n)$ by modular exponentiation and the Euclidean algorithm. If $1<d<n$, it is a proper factor. If a prime divisor $p$ of $n$ has $p-1\mid M$, Fermat's little theorem ensures $p\mid d$; success requires some other prime divisor not to divide $a^M-1$. Common choices are $M=B!$ and $M=\mathrm{lcm}(1,\dots ,B)$. The method exploits a smooth $p-1$, rather than a small factor $p$ itself.

Polynomial division separates a polynomial into a multiple of a chosen divisor and a smaller polynomial remainder. Given $f,g\in F[X]$ over a field $F$, with $g\ne 0$, there are unique polynomials $q,r$ satisfyingHere $q$ is the polynomial quotient. The bound concerns polynomial degree, rather than the sizes of the coefficients.^[ref]

For an original example over the rational numbers, takePolynomial long division first cancels $2X^3$ by subtracting $2Xg$:Subtracting one more copy of $g$ cancels $X^2$:ThusThe final polynomial remainder has polynomial degree one, below the divisor's polynomial degree two. Multiplying out the last identity checks every coefficient, including the constant term $-2+8=6$.

The polynomial degree bound makes this decomposition useful: it replaces a potentially long polynomial with a short polynomial remainder whenever multiples of $g$ can be discarded. At any root of a polynomial of $g$, those multiples vanish under polynomial evaluation. Repeated polynomial division also drives the Euclidean algorithm for finding a greatest common divisor of polynomials.

The coefficient setting matters. Over a commutative ring, the same existence and uniqueness hold when the divisor's leading coefficient is a unit in a ring. A monic polynomial therefore works over the integers, even though they do not form a field; monic polynomial division over a ring explains why zero divisors cause no difficulty in that case.^[ref]

A nonzero divisor alone is insufficient over the integers. Dividing $X$ by $2X+1$ with an integer polynomial quotient and a constant polynomial remainder would require $1=2c$ for an integer $c$, which is impossible. Over the rational numbers the division instead givesThe obstruction is the nonunit leading coefficient $2$. This example separates existence from uniqueness: over an integral domain, the polynomial degree argument still makes any admissible pair unique, if it exists.

A polynomial greatest common divisor of two polynomials over a field divides both and is divisible by every common polynomial divisor. It is unique after requiring it to be monic, and the polynomial Euclidean algorithm computes it.

If $f$ is an irreducible polynomial over a field $F$, then a nonzero residue $[g]$ modulo $(f)$ has $\gcd (f,g)=1$. The Euclidean algorithm supplies $uf+vg=1$, giving inverse $[v]$ for $[g]$. Thus the quotient ring $F[X]/(f)$ is a field. If $\mathrm{deg}f=d$ and $F$ has $q$ elements, polynomial division supplies exactly $q^d$ residue classes.