Solve in set of natural numbers the following systems: \begin{align} &\text{(a)} && x + y = 150,\quad \gcd(x, y) = 30\\[12px] &\text{(b)} && \gcd(x, y) = 45,\quad 7x = 11y\\[12px] &\text{(c)} && xy = 8400,\quad \gcd(x, y) = 20 \end{align}
I've tried Diophantine equations but it did not work.
Maybe there is a clever solution but so far I have been unable to spot it. Can anyone help me or give me any hint?
(a): Since $gcd(x,y) = 30$, then $x = 30x_1$ and $y=30y_1$, where $gcd(x_1, y_1) = 1$. Now substitute in the first equation and get $30(x_1+y_1) = 150$, then $x_1+y_1=5$. It has solutions $(x_1, 5 - x_1)$ where $1 \leq x_1 \leq 4$(all of these pairs are coprime) and $x = 30*x_1$.
(b): $gcd(x, y) = 45$, then $x = 45x_1$ and $y = 45y_1$($x_1$ and $y_1$ - coprime). Then you have $7x=11y$, then $45*7*x_1=11*45*y_1$ which implies that $7$ divides $y_1$ and $11$ divides $x_1$. Notice that every divisor of $x_1$ is also a divisor of $y_1$ except $11$ and every divisor of $y_1$ is divisor of $x_1$ except $7$. But $gcd(x_1, y_1) = 1$, So you must have $x_1 = 11$ and $y_1 = 7$ $\rightarrow$ $x = 45*11$ and $y = 45*7$ for all $x_2, y_2 \in \mathbb{N}$ and $gcd(x_2, y_2) = 1$
Generally when you have problems with $gcd(x, y) = d$ it is a good idea to set $x = d*x_1$ and $y = d * y_1$. I'll leave the third one unsolved(I can solve it if you can, but you can use it as an exercise).