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What are the integers (or at least natural numbers)$x_1$ and $x_2$ which satisfy this equation:

$3x_1 + 7x_2 = 1000$.

And is there a name for such a problem?

Bill Dubuque
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Mira
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    Cf. linear Diophantine equations – J. W. Tanner Mar 08 '21 at 15:44
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    It is worth noting that $\gcd(3,7)=1$ so there exist integers $y_1,y_2$ such that $3y_1+7y_2 = 1$. You can find for instance that $y_1=-2$ and $y_2=1$ works for this. Multiplying by $1000$ gives a solution to the original equation. Further solutions can be found by noting that $3\cdot 7 + 7\cdot (-3)=0$... so the full set of solutions $(x_1,x_2)$ is ${(-2000+7t,1000-3t)~:~t\in\Bbb Z}$ – JMoravitz Mar 08 '21 at 15:49

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This is a linear Diophantine equation.

From the Bezout relation $3(-2)+7(1)=1$, we have $3(-2000)+7(1000)=1000$.

Generally, $3(-2000+7k)+7(1000-3k)=3(-2000)+7(1000)=1000$.

J. W. Tanner
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