Imagine a multiplication table (like the one below, except it continues on forever):

1 2 3 4
+---+---+---+---+---
1| 1 | 2 | 3 | 4 |...
+---+---+---+---+---
2| 2 | 4 | 6 | 8 |...
+---+---+---+---+---
3| 3 | 6 | 9 | 12|...
+---+---+---+---+---
4| 4 | 8 | 12| 16|...
+---+---+---+---+---
|...|...|...|...|...

Find three of the same number in a straight line somewhere within the table. If this is not possible, show why not.

Since the problem asks for NUMBERS and not DIGITS, let's suppose there are three squares, in a straight line, with the same value.

WLOG we can suppose the middle one to be (x,y), one of the others (x+a,y-b), and the third (x-ac,y+bc); a and b must be positive integers, but c could be a rational. Neither a nor b nor c can be zero.

Since xy=(x+a)(y-b), -bx+ay=ab, and since xy=(x-ac)(y+bc), bcx-acy=abc^2. Multiplying the first by c and adding, abc+abc^2=0. Since c isn't zero, c=-1.

Now we need (x+a)(y-b)=(x-a)(y+b), but this leads to ay-bx=bx-ay, so ay=bx, so y=ax/b.

Substituting, we get to a^2=b^2, so a=b, and y=x. But x.x can never be equal to (x+a)(x-a) unless a=0, so the problem is impossible.