Playing with Blocks

G.R.Dixon, 11/11/06

Fig. 1 shows 5^{2} blocks arranged in a square.

Figure 1

5^{2} Blocks

If we remove 4^{2} blocks, we obtain Fig. 2

Figure 2

5^{2} – 4^{2} Blocks

Fig. 3 shows Fig. 2 with 3^{2} blocks added.

Figure 3

5^{2} – 4^{2} + 3^{2} Blocks

Fig. 4 shows Fig. 3 with 2^{2} blocks removed.

Figure 4

5^{2} – 4^{2} + 3^{2} –2^{2} Blocks

Finally, Fig. 5 shows Fig. 4 with 1^{2} blocks added.

Figure 5

5^{2} – 4^{2} + 3^{2} –2^{2} + 1^{2} Blocks

Counting the blocks in the horizontal and vertical rows of Fig. 5, it is clear that

5 + 4 + 3 + 2 + 1 = 5^{2} – 4^{2} + 3^{2} –2^{2} + 1^{2}. (1)

Starting with 4^{2} blocks, it can similarly be shown that

4 + 3 + 2 + 1 = 4^{2} - 3^{2} +2^{2} - 1^{2}. (2)

In general, if N is odd then

N + (N – 1) + (N – 2) + … + 1 = N^{2} – (N – 1)^{2} … + 1. (3)

And if N is even then

N + (N – 1) + (N – 2) + … + 1 = N^{2} – (N – 1)^{2} … - 1. (4)

Note that, as N grows without bound, Fig. 5 approaches a right triangle of area N^{2} / 2. Whence we conclude that in the limit of arbitrarily large N,

N^{2} – (N – 1)^{2} … __+__ 1 = N^{2} / 2. (5)

Or, equivalently, in the limit,

N + (N – 1) + … + 1 = N^{2} / 2. (6)

It is readily demonstrated that, for certain integer values of N and M, with N > M, N^{2} - M^{2} produces integer squares. For example, in Fig. 2, 5^{2} – 4^{2} results in 9 = 3^{2} blocks. More generally, certain integer values of N and a, with N > a and M = N - a, produce perfect squares for N^{2} – (N - a)^{2}. Other combinations of N and a do not produce such a perfect square result.

It would also seem that exponents greater than 2 cannot produce any "perfect" results at all … a proposition first suggested by Fermat. For example, if M = N - a, then

N^{3} – M^{3} = 3aN(N - a) - a^{3}.
(7)

According to Fermat, there are no integers a and N such that the right side of Eq. 7 has an integer cube root. Similar remarks presumably apply to higher integer exponents. In general Fermat claimed to have a simple proof that there is no integer I, such that

N^{n} – M^{n} = I^{n}, n an integer >2 and N, M integers. (8)

His proof was either never written down, or if so was never found. Countless doodlers have tried since his death to figure out what he might have had in mind.