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Find the number of points on the straight line which joins $(-4,11)$ and $(16,-1)$ whose coordinates are positive integers.

a) $1$

b) $2$

c) $3$

d) $4$

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The slope of this line is $$\frac{-1-11}{16-(-4)}=\frac{-12}{20}=-\frac35\;.$$ Thus, the equation of the line is $y-11=-\frac35(x-(-4))=-\frac35x-\frac{12}5=-\frac15(3x+12)$. Clearly $y$ is an integer precisely when $3x+12$ is a multiple of $5$, which is the case precisely when $x+4$ is a multiple of $5$. These values of $x$ occur $5$ units apart, so between $-4$ and $16$ they are $-4,1,6,11$, and $16$, and the corresponding $y$-coordinates are $11,8,5,2$, and $-1$.

You don’t actually need the equation of the line. The slope of $-\frac35$ means that every $1$-unit increase in $x$ causes a decrease in $y$ of $\frac35$. Thus, it takes a $5$-unit increase in $x$ to decrease $y$ by an integer (namely, $3$), and the values of $x$ between $-4$ and $16$ that give integral values of $y$ must be $-4,1,6,11$, and $16$. The corresponding $y$-coordinates are $11,8,5,2$, and $-1$.

By either approach we see that the line segment contains exactly $3$ points whose coordinates are both positive integers.

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The equation of the line is $y-11=(\frac{11+1}{-4-16})(x+4)=-\frac{3}{5}(x+4)$. This means $3x+5y=43$.

Clearly $(1,8)$ is a point on the line such that $x$ and $y$ are both positive integers, and other such points are found by increasing $x$ by $5$ and decreasing $y$ by $3$. We then see that the only other such points are $(6,5)$ and $(11,2)$.

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How did you find these solutions? Why aren't there more? This is an incomplete answer. – akkkk Dec 20 '12 at 9:42
@akkkk: There’s nothing wrong with an incomplete answer that contains enough information to allow the details to be filled in reasonably easily, which was the case with Jasper’s original answer. – Brian M. Scott Dec 20 '12 at 9:48
@akkkk: To find the first solution in cases like this, one can use Bezout's lemma (aka Bezout's identity). – Shane O Rourke Dec 20 '12 at 9:52
@ShaneORourke: oh don't worry, I know how to find the answer, but clearly something's missing. – akkkk Dec 20 '12 at 11:16

The straight segment of line having those points as its extremes is

$$t(16,-1)+(1-t)(-4,11)=(20t-4\,,\,-12t+11)\,\,\,,\,\,0\leq t\leq 1$$

Well, it is plain that it must be that both $\,20t\,,\,12t\in\Bbb Z\,\,\,,\,\,t\in [0,1]$ , so

1) Prove that it must be $\,t\in\Bbb Q\cap [0,1]\,$

2) It also must be that, if $\,t=m/n\,$ , then $\,n\,\mid 20\,\,\,\text{and also }\,\,\,n\,\mid 12\,$ or else $\,m=0\,$

BTW, if we count the extreme points (and I don't understand why wouldn't we) , then the correct answer doesn't show up in the possible answers...

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We don't care for the extreme points because last time I checked, $-4$ and $-1$ failed to be positive – Hagen von Eitzen Dec 20 '12 at 9:36

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