GCD of two real numbers

Question

How would I show that gcd($2a+1 , 9a+4)=1 $? Here $a$ is an integer. I used the definition of the greatest common divisor, but felt it is too lengthy.

Answer 1

Observe that $\displaystyle9(2a+1)-2(9a+4)=1$

So, if real number $d$ divides both $2a+1,9a+4;$

$d$ will divide $9(2a+1)-2(9a+4)=1$

Answer 2

The fastest method I know of is Euclid's algorithm. We have:

$$GCD(2a + 1, 9a + 4) = GCD(2a + 1, a) = GCD(1, a) = 1$$

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Here is an alternative albeit slower method, in case Euclid's algorithm is not intuitive enough.

Let $d$ be the gcd of $2a + 1$ and $9a + 4$. Then,

$$d \mid 2a + 1 \implies d \mid 18a + 9$$

Similarly,

$$d\mid9a + 4 \implies d \mid 18a + 8$$

But if $d \mid m$ and $d\mid n$, then $d \mid m-n$. Hence,

$$d \mid (18a + 9) - (18a + 8) = 1 \implies d \mid 1$$

The only way is if $d = 1$. Hence, gcd of the two is equal to $1$.