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Consider the equation $ax + by = (a+b)z$ with $a < z$ and $b > z$. The variables a, b, z are fixed and one can vary x and y. Why is there always a solution to this equation? It might be obvious but I somehow don't see it... I maybe should also add that $a, b, z, a, x, y \in \mathbb{N}$

Hilberto1
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  • An integer solution for $x$ and $y$ for fixed, real $a,b,z$? You also noted that there was claimed to always be an integer solution and then claimed that $a,b,z,x,y \in \mathbb{N}$, which sort of solves the question by assumption. – Alex Hiller Jan 18 '21 at 21:13
  • I corrected it. A solution in N where all involved variables are also in N – Hilberto1 Jan 18 '21 at 21:23
  • @ThéodorLemerle Your comment no longer is appropriate. – amWhy Jan 18 '21 at 21:43
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    Since $\ az+bz=(a+b)z\ $, then $\ x=y=z\ $ is always a solution of $\ ax+by=(a+b)z\ $. Is there any reason why this solution is not satisfactory? – lonza leggiera Jan 18 '21 at 22:52
  • You can also do $ax+by=(a+b)z\implies by = bz + a(z-x)$ Long as $\frac b{\gcd(a,b)}|z-x$ we are good. And as $x$ can be anything we can always do that. (and letting $z = x$ is the easiest way to do that. – fleablood Jan 18 '21 at 23:30

1 Answers1

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Suppose $a = 0$.

Then you need to show that

$0*x + by = (0+b)z$ has a solution. It obviously does if $x = anything$ and $y = z$.

Then $0*anything + bz = (0 + b)z$. Which is true.

Suppose $b = 0$ then we need to show that

$ax + 0*y = (a+0)z$ has a solution. It obviously does if $x = z$ and $y = anything$.

The $az + 0*anything =(a + 0)z$. Which is true.

And if $a\ne 0; b\ne 0$ then

$ax + by = (a+b)z$

$ax = (a+b)z-by$

$x = \frac {(a+b)z- by}a = z + \frac ba(z -y)$

If $a$ divides into $b$ evenly then $\frac ba$ is an integer.

If we let $y = anything$ and $x = z + \frac ba(z-y)$ we have a solution.

$a(z + \frac ba(z-y))+by=$
$(az + bz -by)+by = az+bz = (a+b) z$.

Otherwise if $\frac ba$ reduces to the fraction $\frac {b'}{a'}$ (which means $a'b=ab'$).

Then if we let $r = $ the remainder of $z$ when divided by $a'$ (so that there is a $q$ so that $z = qa' +r$) then if we let $y = ka'+r$ for any $k$ and let $x = z +b'q -b'rk$

Then we have a solution

$ax + by = $
$a(z +b'q -b'rk) +b(ka' + r)=$
$az + ab'q - ab'rk + a'bk + br=$
$az + a'bq+br = az + b(a'q+r) =az + bz = (a+b)z$

fleablood
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