When we learn about unsolved problems, no matter the discipline, we are generally indoctrinated into the patterns of thought that haven’t been able to bring about a solution. In many cases, these patterns will eventually yield an answer. For instance, the periodic table turned out to be a good guide for thinking nebulium wasn’t a real element. But in other cases, the ideas of our predecessors can actually be a hindrance. To illustrate, if a community of doctors believed that bloodletting were a curative for fevers, the question, “How do we cure scarlet fever?” might be incorrectly narrowed to the question, “How many leeches are necessary for removing scarlet fever from the blood?” Experiments varying the number of leeches used would never produce the cure for scarlet fever, but it isn’t difficult to imagine an empirical experiment showing that 4 leeches are better for the patient’s health than 20. It’s even imaginable that one leech might be  “shown” to be more beneficial to the patient than none—perhaps the presence of one leech leads to patients’ performing better on whatever indicators of health the scientist has chosen for the experiment (e.g. the leech lessens the patient’s flushed appearance).

It isn’t always easy to shake off the old paradigm of thinking—the more we learn on a topic, the deeper we sometimes settle into its ruts. And while much superstition says 30 is over-the-hill for mathematicians (here’s a piece that challenges the stereotype), it’s possible that mathematicians settle into the status quo through exposure more than age. Even if you’ve never thought much about number theory, or especially if you’ve never thought much about number theory, and if you consider yourself to be a puzzle solver or a problem solver in everyday life, now is your chance to dive in and shake things up. So I’m here to give you some tools to do it. Enjoy!

A successful problem solver is someone who figures things out. That person could have a knack for solving problems, or she could have a drive to do it (or both). If solutions don’t come easily to you, don’t let this exclude you from the ranks of problem solvers in the world. Solutions to great problems have been found because their solvers just wouldn’t give up. But this is starting to sound like a motivational speech, so let’s get down to business. I want you to prove Goldbach’s as-yet unproven conjecture about even numbers. I am going to provide you with two examples of mathematical proof that are comprehensible in laymen’s terms, and then I’ll lay out the problem. When you figure it out, contact your nearest mathematician to secure bountiful rewards.

Puzzle Walkthrough 1: The Mutilated Chessboard

Begin with an empty chessboard. A chessboard consists of 64 squares that alternate between light and dark, so that no light square is ever adjacent to another light square (and vice versa), arranged in 8 rows of 8 squares to create a large square. Now suppose you remove two diagonal corner squares from the board, leaving the remainder of the board intact. This entails removing either two light squares or two dark squares, and leaving 62 squares as they are. You end up with either 30 light squares and 32 dark squares or 30 dark squares and 32 light squares, depending on which two corners were removed. Suppose you also have 31 dominoes. Each domino is the size and shape of two adjacent squares on your chessboard. It is possible to place a domino on the board so that it completely covers two adjacent squares, and only covers those two adjacent squares. The question is, can you place your 31 dominoes in such a way that they completely cover every square on the board?

And here’s the proof that you can’t: Because no light square can be adjacent to another light square and no dark square can be adjacent to another dark square, we know that one domino cannot cover two light squares or two dark squares: every domino is capable of covering one square and one of its adjacent squares (or one light square and one dark square). 31 dominoes, then, can together cover 31 dark squares and 31 light squares. If our board has 62 squares covered, 31 of them must be light and 31 dark. But wait! We removed two like-colored squares, leaving 32 squares of one color and 30 squares of the other. We can therefore cover 30 pairs of squares with dominoes, but we’ll be left with two remaining non-adjacent squares. To cover these, we would have to mutilate a domino too, which isn’t an option. We therefore know that we cannot cover all the squares of our mutilated chessboard with 31 dominoes.[1]

Puzzle Walkthrough 2: Euclid’s Proof of the Infinitude of Primes

Starting with any prime number, assume we have found all the prime numbers. For our example, we’ll assume that 7 is the highest prime number, and we have found all the prime numbers through 7. We have 2, 3, 5, and 7 (note that 1 is not a prime number). The product of these primes will be divisible by each of them. 2 × 3 × 5 × 7 = 210. Now clearly we can divide 210 by any of these numbers and get an integer, because we multiplied by all of these numbers to begin with (in the same way, when we multiply 2 and 3 to get 6 we can divide 6 by 2 to get 3 or by 3 to get 2). But if we take the product of our primes and add 1 to it (so, [(2 × 3 × 5 × 7) + 1 = 211]), suddenly our number is not divisible by any of the primes we have.

This is easy to see with a bit of thought. If some number is divisible by 2 and we add 1 to the number, it will no longer be divisible by 2, because it will give us a remainder of 1. A new number divisible by 2 would be obtained by adding 2 to the original number. Consider 6 again. We know that 6 is the product of 2 and 3, so we know that 6 is divisible by 2. If we add 1 to 6, when we divide 6 by 2, we get a remainder of 1. If we add 2 to 6 to get 8, then we have another number divisible by 2. In the same way, by adding 1 to our product of primes, we create a number that is not divisible by any of them.

Every natural number is either prime or it is not prime: either we can divide it by some integer (other than 1 and itself) to obtain another integer, or we cannot. Likewise, [(2 × 3 × 5 × 7) + 1] is either prime or it is not. If it is prime, we have a new prime number. The number 211 does happen to be prime. But if it were not prime, it would have to be factorable into numbers that are prime. That is, if we divide a divisible number, we’ll reach a stopping place where the integers we end up with are themselves indivisible. For example, if we have the number 16, which is divisible by the prime number 2, and we divide it, we get 8. This is again divisible by 2, leaving us 4. If we divide 4 by 2, we get 2: thus 16 can be expressed in terms of prime constituents 2, 2, 2, and 2. We know that our product 211 isn’t divisible by 2, 3, 5, or 7. But supposing we didn’t know that it is prime, let’s consider what it would mean for it to be divisible. It would mean that dividing it, we would reach a stopping place of constituent primes. Because we know that these constituents can’t be 2, 3, 5, or 7, we also know they would have to be another number that is prime.

Thus the product of these primes + 1 gives us a new prime number either by being itself prime or by entailing a factor that is prime and not one of the primes we already considered. We can easily generalize this proof to all known primes: one plus the product of the set of all known prime numbers will yield a number that is either prime or divisible by some prime number that was not included in the set being multiplied. Euclid was a genius.

Euclid’s proof is a great illustration of how abstract reasoning can trump raw calculation. No matter how many new primes we discover, the fact that we have discovered more does not prove that more are discoverable,[2] just as the fact that the sun has risen perhaps more than a trillion times does not prove that it will rise tomorrow. It makes it very, very statistically likely, but it does not show us a necessary truth. Statistical probability can provide the likelihood that a certain series of numbers will contain a prime number, based on the size of the numbers in the sequence and the observation that the frequency of the appearance of primes tends to decrease as the size of the numbers in the sequence increases (Marcus du Sautoy gives good pop-explanations of this topic). But observing the distribution of primes among the numbers we know does not tell us whether the appearance of primes is always approaching zero without ever reaching it (asymptotically), or whether we can find a number large enough that all numbers larger than it will not be prime (will be divisible by some number other than themselves and 1). Proof tells us, and proof is easier.

The Goldbach Conjecture

Any even number except the number 2 can be expressed as the sum of two primes.[3]

This has been shown empirically for all even numbers from 4 to 4 x 10¹⁸, but it’s never been proven. Thus there is a nonzero possibility (no matter how miniscule) that we’ll get to, say, 4 x 10²⁸ and find an even number for which Goldbach’s conjecture doesn’t hold.[4] Although we can say with some practical confidence that the conjecture is likely true for all even numbers, we can’t say it with certainty unless we find a reason that it must be the case. And if it is in fact true for all even numbers, there is certainly a reason.[5] Any of the even numbers in question will be expressible as the sum of a prime number and some other number, because any even number n will contain every prime number less than n. For instance, the number 11 can be expressed as 2 + 9, 3 + 8, 5 + 6, 7 + 4, and even 11 + 0, with the first number in each pair being prime. But why should it be that any even number contains two prime numbers which when added together produce that number? Yet the evidence we have suggests this is the case. 8 = 3 + 5, 10 = 5 + 5 or 7 + 3, and so on.

If it were possible to calculate all possible sums of every even integer greater than 2, then by doing this we could discover that Goldbach’s conjecture is true for all of these numbers (assuming it is true). This method would be sufficient to establish the truth of the conjecture, though it wouldn’t be very efficient. But, again, we can’t possibly take this route, because the numbers are infinite, so whatever we establish about the numbers we crunch cannot predict the behavior of the numbers we haven’t crunched, meaning in order to prove the conjecture we must find a way to demonstrate why it must be the case for any even number greater than 2 that it is the sum of two prime numbers.

“The source of the difficulty [in proving Goldbach’s conjecture] is that primes are defined in terms of multiplication, while the problem involves addition. Generally speaking, it is difficult to establish connections between the multiplicative and the additive properties of integers.”[6] This is not for lack of trying. The history of progress on the problem is not really our concern here (but here’s a link), because I want to focus on the unsullied perspective non-mathematicians might have to offer. Complex mathematics aside, if even numbers all share some property that allows them to be expressed as two primes—or perhaps if prime numbers are distributed such that if every prime number is added to every other prime number in pairs (excluding 2 from the group, as 2 + an odd prime would create an odd number) (e.g., 3 + 5, 3 + 7 … 3 +n, 5 + 5, 5 + 7 … 5 + n, and so on), the sums will include every even number—then looking at the numbers in just the right way could conceivably give any one of us non-mathematicians the key to unlocking Goldbach’s conjecture. While it seems natural for mathematicians to continue whittling away at the work that’s already done, the beauty of approaching the problem from the outside is that we have the potential to cut new paths in our initial approach to the problem (though obviously we could still end up retracing the old ones!).

Proving Goldbach’s conjecture could lead to useful knowledge about the distribution of primes. And knowledge about the distribution of primes has real-world applications. This might[7] be a very basic example: If an even number n can be expressed as the sum of two primes, either the primes are identical (e.g., the prime 5 doubled equals the even number 10) or one of the primes must be greater than half of n. If this is true for any even number, then no even number can exist inside a prime gap that extends beyond half of that number (e.g. the even number 30 must contain a prime that is greater than or equal to 15 and less than 30), meaning a prime gap immediately preceding any arbitrarily chosen even number could not be more than half that number. But even supposing proving the conjecture had no useful application at all, imagine how satisfying it would be to make waves in the mathematical world as an outsider.

We manipulate numbers every day just by negotiating reality, whether we put our pants on one leg at a time or two (or go naked). The ubiquity of numbers alone means none of us approaches number theory entirely without experience, and our lack of experience could be just the source of brilliance necessary to solve the problem. Why not try? After you prove the Goldbach conjecture, you can make some fast cash by solving the millennium problems, or solve the problems on this list (to achieve what I can only assume would be some sort of nirvana). Happy proving.