I am having a hard time with this problem and I would really like some help
Let $D=\gcd(a,b)$, $a = \alpha'\times D$ and $b = \beta' \times D$, all nonzero naturals.
Find the possible $D,\alpha,\beta$ that satisfies $D(3\alpha' + D\beta')=3388$. $D$ must be prime and $b \gt 49$
What I've done is decomposing $3388=2^4*7*11^2$, so $D \in \{2,7,11\}$
But now I have to do what $\alpha',\beta'$ is, for each $D$, how can I do this? Do I have to give $b$ a value grater than 49 and try each time? That seems unefficient
Suppose $D = 11$
$\frac {3388}{D} = 308\\ \gcd(3,11) = 1\\ 3\cdot(4) + 11\cdot(-1) = 1\\ 3\cdot(4\cdot 308) + 11\cdot(-308) = 308\\ 3\cdot(4\cdot 308+11m) + 11\cdot(-308+3n) = 308\\ 11(3\cdot(4\cdot 308-11k) + 11\cdot(-308+3k)) = 3388\\ $
Find $k$ that fit our constraints
$4\cdot 308 - 11k > 0\\ -308 + 3k \ge 49$
$k<112\\ k>119$
We don't have consistent solutions.
$D = 7\\ 3\cdot(-2) + 7\cdot(1) = 1\\ 7(3\cdot((-2)\cdot 484-7k) + 7\cdot(484+3k)) = 3388\\ -968-7k > 0\\ 484 + 3k > 49\\ -161\le k\le -145$
And I will leave it to you to work out $D = 2$