Suppose $\gcd(50+x^2,4x+2)=6$ What is the smallest positive integer that $x$ could be?
It seems that this could be solved with modular arithmetic, however I would be interested to see if there would be another way which wouldn't depend on using modular arithmetic. Any suggestions would be appreciated!
Since $3 \mid 4 x + 2$, $x \equiv 1 \mod 3$. Since $2 \mid 50 + x^2$, $x$ is even. The first even number $\equiv 1 \mod 3$ is $4$. And that turns out to work.
Hint $\,\ 6=(a,b)=(a,b,6) = (a,b,2)(a,b,3) = 2\cdot 3$
Here $\ (50\!+\!x^2,4x\!+\!2,2)=(x^2,\,2)\ =\ 2\ \,\iff\:\!\ 2\mid x,\ $ and
and $\ \ \,(50\!+\!x^2,4x\!+\!2,3^{\phantom{|^{|^|}}}\!\!\!\!) = \underbrace{(x^2\!-\!1,x\!-\!1,3)}_{\textstyle (x\!-\!1,3)}\!\!\iff\! 3\mid x\!-\!1$
using $\,(a,b,n) = (a\bmod n,\, b\bmod n,\,n) =$ Euclidean gcd reduction (for $\,x\!-\!1\mid x^2\!-\!1$ too)
if $6\big|(x^2+50)(4x+2)$ then:
$$24\big|(4x^2+200)(4x+2)$$
$$4x^2+200=x(4x+2)+200-2x$$
$(200-2x)$ is the remainder of division of $4x^2+200$ by $(4x+2)$ and must be divisible by 24(or we may say $6|(x^2+50)(4x+2)$ can be reduced to $24|(200-2x) \ $), so we must have :
$$ 200-2x=24k \implies x=100-12k$$
To minimize $x$, we maximize $k$:
$k=9 \implies x=100-108=-8$ is not acceptable.
$k=8 \implies x=100-96=4$ is acceptable.
