Or even just a *math smarter-than-I-am?* Here’s your chance to prove it…

Test your math and problem solving skills with this brain teaser. No cheating! 😉

**The Dutchmen’s Wives**

Three Dutchmen, named Hendrick, Elas, and Cornelius, and their wives, Gurtrün, Katrün, and Anna, purchase hogs.

Each buys as many as he (or she) gives shillings for one.

Each husband pays altogether three guineas more than his wife. Hendrick buys 23 more hogs than Katrün, and Elas eleven more than Gurtrün.

Now, what was the name of each man’s wife?

Scroll down for the answers…

**Here is the Answer:**

The money paid in every case was a square number of shillings, because they bought 1 at 1*s*., 2 at 2*s*., 3 at 3*s*., and so on.

But every husband pays altogether 63*s*. more than his wife, so we have to find in how many ways 63 may be the difference between two square numbers.

These are the three only possible ways: the square of 8 less the square of 1, the square of 12 less the square of 9, and the square of 32 less the square of 31.

Here 1, 9, and 31 represent the number of pigs bought and the number of shillings per pig paid by each woman, and 8, 12, and 32 the same in the case of their respective husbands.

From the further information given as to their purchases, we can now pair them off as follows:

**Cornelius and Gurtrün bought 8 and 1; Elas and Katrün bought 12 and 9; Hendrick and Anna bought 32 and 31.**

** **

**And these pairs represent correctly the three married couples.**

The reader may here desire to know how we may determine the maximum number of ways in which a number may be expressed as the difference between two squares, and how we are to find the actual squares.

Any integer except 1, 4, and twice any odd number, may be expressed as the difference of two integral squares in as many ways as it can be split up into pairs of factors, counting 1 as a factor.

Suppose the number to be 5,940.

The factors are 2^{2}.3^{3}.5.11. Here the exponents are 2, 3, 1, 1.

Always deduct 1 from the exponents of 2 and add 1 to all the other exponents; then we get 1, 4, 2, 2, and half the product of these four numbers will be the required number of ways in which 5,940 may be the difference of two squares—that is, 8.

To find these eight squares, as it is an even number, we first divide by 4 and get 1485, the eight pairs of factors of which are 1 × 1485, 3 × 495, 5 × 297, 9 × 165, 11 × 135, 15 × 99, 27 × 55, and 33 × 45.

The sum and difference of any one of these pairs will give the required numbers.

Thus, the square of 1,486 less the square of 1,484 is 5,940, the square of 498 less the square of 492 is the same, and so on.

In the case of 63 above, the number is odd; so we factorize at once, 1 × 63, 3 × 21, 7 × 9.

Then we find that half the sum and difference will give us the numbers 32 and 31, 12 and 9, and 8 and 1, as shown in the solution to the puzzle.

The reverse problem, to find the factors of a number when you have expressed it as the difference of two squares, is obvious.

For example, the sum and difference of any pair of numbers in the last sentence will give us the factors of 63.

Every prime number (except 1 and 2) may be expressed as the difference of two squares in one way, and in one way only.

If a number can be expressed as the difference of two squares in more than one way, it is composite; and having so expressed it, we may at once obtain the factors, as we have seen.

Fermat showed in a letter to Mersenne or Frénicle, in 1643, how we may discover whether a number may be expressed as the difference of two squares in more than one way, or proved to be a prime.

But the method, when dealing with large numbers, is necessarily tedious, though in practice it may be considerably shortened.

In many cases it is the shortest method known for factorizing large numbers, and I have always held the opinion that Fermat used it in performing a certain feat in factorizing that is historical and wrapped in mystery.

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