Is there a way how to determine number of solutions in linear equation like this: $ax + by + cz = d$, where $a,b,c,x,y,z,d$ are non-negative integers and $a,b,c,d$ are known?
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2You can use generating functions and look at the coefficient of $X^d$ in the expansion of $(1+X^a+X^{2a}+X^{3a}+\dots)(1+X^b+X^{2b}+\dots)(1+X^c+X^{2c}+\dots)$ – JMoravitz Nov 14 '18 at 19:50
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See also this question and similar ones. – Dietrich Burde Nov 14 '18 at 19:57
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Above equation shown below:
$ax + by + cz = d$ -------$(1)$
For (a, b, c, d)= (3, 2, 5, 32), equation $(1)$ has parametric solution given below:
$x=(1-v)$
$y=(-5u+9v+87)$
$z=(2u-3v-29)$
For $(u,v)$ = $(15, 0)$ we get,
$(x, y, z)$ = $(1, 12, 1)$
Sam
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