Deciding parity without computing the answer

0:00Write four numbers in a row. Three, four, five, six. Between them there are three gaps, and into each gap you can put a plus or a minus. Two choices, three times over. Two times two times two. Eight expressions, all built from the same four numbers. Work all eight of them out, and something happens that has no business being obvious in advance. Every single one of the eight lands on an even number.

0:30Not usually. Every time. And that is worth explaining rather than admiring. Here is how to be sure that none of the eight has been missed. Start at three. The first gap splits into two branches: plus four, and minus four. Each of those splits again at five, and each of those splits again at six. One level per gap, doubling each time. Two, then four, then eight leaves. Every path down the tree spells out one expression, and every expression is one path.

1:06So there are exactly eight of them, and the tree is the reason we know that. No hunting, and no wondering whether a ninth is hiding somewhere. Take a couple of them and work them out. Three plus four minus five plus six is eight. Three minus four minus five minus six is minus twelve. Both even, and one of them negative, which is worth pausing on, because it is where half of this topic gets thrown away.

1:38Minus twelve is even. So are minus two, minus four and minus six. An even number is one with two as a factor, and the sign has nothing to do with it. Start from five, six, seven, eight instead and you get twelve, and minus sixteen. Even again. Now the real question. Why can none of the eight come out odd? Take the all-plus one. Three plus four plus five plus six is eighteen.

2:09Change a single sign. The plus in front of the four becomes a minus. The new total is ten. Eighteen to ten. It has moved by eight. And eight is twice four, which is exactly the term whose sign we touched. That is not a coincidence about this run of numbers. It happens every time. Here is why. The four was being added on; now it is being taken away.

2:39So the total has lost that four twice over. Once for the adding that stopped, once for the taking away that started. Switch a minus back to a plus and the same thing runs the other way. The total gains twice that term. Either direction, the move is twice something. And twice anything is even. Adding or taking away an even amount cannot carry a number across from even to odd.

3:08So a sign switch changes the total and never changes its parity. That is the step the whole topic rests on. Checked on a hundred and sixty-four sets of four numbers, then again on two hundred and sixty-four more, at every gap of every signing: the move is twice that term, and it is even. One switch keeps the parity. But there are eight expressions here, not two. So walk between them. From the all-plus one, switch the first gap.

3:43Switch the second. Switch the third. Switch one of them back. Every one of the eight can be reached from every other, one switch at a time. Three switches is the furthest apart any two of them get, which is one per gap. And every step along the way keeps the parity. So all eight share the parity of the one we started from. The plain total is eighteen, and eighteen is even, so all eight are even.

4:13And the plain total is the one value you can work out without choosing any signs at all. There is a second way to see this, and it never mentions switching. Odd with odd gives even. Even with even gives even. Odd with even gives odd. And each of those three covers a difference as well as a sum. Seven plus five is twelve. Seven minus five is two. Both even.

4:42Because taking a number away and adding it on shift the total by the same amount. So chain the rules along the expression one term at a time, and the signs simply never come into it. Each step only ever asks whether the next term is odd or even. Three, four, five, six. Odd with even is odd; that with odd is even; that with even is even. And a third way, with no symbols in it at all.

5:14Picture each number as a pile of counters. A plus makes it a positive pile, a minus a negative one. A positive counter and a negative counter cancel, and both of them go. They always leave in pairs. Two at a time, never one. So whatever the signs are, the number of counters that disappears is even. What is left over is the answer, and its parity was settled before any cancelling started.

5:45Three ways of saying the same thing, and not one of them ever needed to know what the four numbers actually were. Which raises the obvious question. Did any of that mention four? It did not. Two numbers have one gap and two expressions. Five numbers have four gaps and sixteen. Seven numbers have six gaps and sixty-four expressions. And in every case a sign switch moves the total by twice a term, so every expression from one block shares one parity.

6:21Checked across a hundred and eight blocks, of every length from two up to seven. And the side they land on is the parity of the plain total, which you can know without writing down a single expression. So parity comes off the shape of an expression rather than its value, which means you can answer a question without doing the arithmetic in it. Here are eight, and none of them is to be worked out.

6:52Forty-three plus thirty-seven. Odd with odd. Even. Six hundred and seventy-two minus three hundred and forty-eight. Even with even. Even. Seven hundred and eight minus four hundred and seventy-seven. Even with odd. Odd. Products behave differently. A product is odd only when every single factor is odd. So four times three hundred and forty-seven times three is even because of that four alone, and a hundred and nineteen times three hundred and three is odd.

7:29Four of the eight come out even and four odd, and not one of them was multiplied out. Now the same question with letters in it, where even has to mean even for every integer at once. Two a plus two b. Both parts are even whatever a and b are. Always even. Three g plus five h. Two odd coefficients, but that is not how coefficients work. Three g carries the parity of g, five h carries the parity of h, and odd with even is odd. So it depends.

8:06Thirteen k minus five k looks odd for the same reason and is not. It simplifies to eight k. Six m minus three n has a six on the front and is still not always even, because the three hands the parity straight to n. Of the nine, five are even for every integer and four are not. And the four that fail are not never even. Take x squared plus two.

8:36Put in six and you get thirty-eight. Even. Does that settle it? Put in three and you get eleven. Odd. One value can destroy a claim about every integer. No value can establish one, and no number of values can either, because there is always another integer you have not tried. When a claim is true there are two ways to show it. Four m plus two q: both of its parts are even.

9:06Or: it is two times, two m plus q. A two lifted clean out of the whole thing. Put in m of four and q of minus nine and it comes to minus two. Those are not the same argument. Six m plus two n is two times three m plus n whether or not three m is even, so the factor survives exactly where the parts do not.