I want to show that there are no integers $x, y$ that solve the following equation:

$$ 15x-9y=100 $$

My solution would be:

$$ 3\cdot5\cdot x-3\cdot3\cdot y = 100 \\ 3\cdot(5x-3y) = 100 \\ 5x-3y = \frac{100}{3} $$

Which means that $x$ and $y$ can't be integers. Is this correct?

  • $\begingroup$ Yes, it is perfectly ok. $\endgroup$ Jul 22, 2015 at 16:00

1 Answer 1


Yes, this is correct.

A slightly shorter argument would be that 3 is a divisor of the LHS because $15x-9y=3(5x-3y)$ and it is clearly not a divisor of the RHS.

  • 1
    $\begingroup$ That's basically the same thing he's saying, though, just in more concise terms $\endgroup$
    – Stephen
    Jul 22, 2015 at 15:41

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