11 votes

You're going to laugh, but the nonzero integers have a natural group law using multiplication and gcd formula (no addition!)

Unfortunately, $4 \ast 2=\frac{8}{4}=2=1 \ast 2$, so $\ast$ can’t be a group law.
Aphelli's user avatar
  • 31.7k
6 votes

Polynomial gcd of $x^5+x+1$ and $x^2+1$

Let $d(x)$ be the greatest common divisor, so that $d(x)\mid x^2+1$ and $d(x)\mid x^5+x+1$. Suppose $d(x)\ne 1$. Then any root of $d$ is a root of $x^2+1$, so equals $\pm i$. Any root of $d$ is a root ...
ancient mathematician's user avatar
3 votes

Possible values of $\gcd(2a^2+6a-4, 2a^2+4a-3)$

Any common factor of $2a^2+6a-4$ and $2a^2+4a-3$ is also a factor of $(2a^2+6a-4)-(2a^2+4a-3)=2a-1$ and hence of $(2a^2+4a-3)-a(2a-1)=5a-3$ and hence of $5(2a-1)-2(5a-3)=1$. If you wish you can ...
David's user avatar
  • 80.3k
2 votes

Polynomial gcd of $x^5+x+1$ and $x^2+1$

One other way I see is to factorize $x^5+x+1$ over $\Bbb Q$. Since it has no rational root by the rational root theorem, we may assume that it is a product of a quadratic and a cubic polynomial. This ...
Dietrich Burde's user avatar
1 vote

Polynomial gcd of $x^5+x+1$ and $x^2+1$

Go to complex numbers, as the GCD is the same over real numbers and over complex numbers. (Euclid's algorithm doesn't "care" if you are executing it in $\mathbb R$ or in a bigger field.) So, ...
Stinking Bishop's user avatar

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