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(From I. M. Gelfand) If it is known that the straight line y = kx + b passes through two integral points, are there any other integral point on this straight line?

I tried out contradiction. Suppose (x1, y1) and (x2, y2) are the two integral points through which the given line passes. And further, that there are no more integral points on this line. So, any other point, say (x3, y3) on the line cannot be integral.

I have three equations: y1 = kx1 + b; y2 = kx2 + b; y3 = kx3 + b

and can get to a relation between (x1, y1), (x2, y2) and (x3, y3) but this looks like a long drawn process. Any Hint?

buzaku
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change $x$ by $x_2-x_1$ changes the value of $y$ by an integer so just add $x_2-x_1$ to $x_2$ to get another integer. There are therefore an infinite number of integral points.

Mark Joshi
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  • It is not mentioned that k is an integer. For instance, the line y = (7x/15) + (1/3) passes through (10,5) which is an integral point. Following your logic though, we can prove that a linear combination of the type (nx1 + (1-n)x2, ny1 + (1-n)y2) also satisfies the equation. Hence, there are infinite solutions. Hope there is still a more elegant solution! – buzaku Apr 05 '15 at 10:54
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    $k$ doesn't need to be an integer: just that $k(x_2-x_1)$ is and this equals $y_2-y_1$ which is one. – Mark Joshi Apr 05 '15 at 10:56