# Curious Identities Involving Integer Squares

One characteristic feature of the identities below is the inclusion of integers with long sequences of equal successive digits. To make describing such numbers a manageable undertaking, I'll use subscripts to denote the number of repetitions of a digit, e.g., 1532 will denote 1111133, i.e., five 1s followed by two 3s. In general, dn means a sequence of n d's.

The following identities come from the Mathematics Magazine, Vol. 35, No. 2 (March-April, 1962). For k > 1,

• 1k5k-16 = (3k-14)², e.g., 11115556 = 3334². (Proof)

• 1k·10k-15 + 1 = (3k-14)², e.g., 1111·10005 + 1 = 3334². (Proof)

• 9k-1820k-181 = (9k1)², e.g., 999982000081 = 999991². (Proof)

• 9k-1840k-164 = (9k2)², e.g, 9998400064 = 99992².

• 9k-1860k-149 = (9k3)², e.g, 9998600049 = 99993².

• 9k-1880k-136 = (9k4)², e.g, 9998800036 = 99994².

• 9k0k25 = (9k5)², e.g, 9999000025 = 99995².

• 5k-16² - 4k-15² = 12k, e.g., 556² - 445² = 111111. (Proof)

Subscripts can be easily converted to decimal notations. Observe that

1t = 9t / 9 = (10t - 1) / 9.

So, for example,

3k5m6n7 = 7 + 10·6·(10n - 1) / 9 + 10n + 1·5·(10m - 1) / 9 + 10m + n + 1·3·(10k - 1) / 9.

### Number Curiosities

1k5k-16 = (3k-14)².

 1k5k-16 = 6 + 50·(10k-1 - 1) / 9 + 10k (10k - 1)/9 = 6 + 5·10k / 9 + 102k / 9 + 6 - 50/9 - 10k/9 = (4·10k + 102k + 4)/9 = [(10k + 2) / 3]².

1k·10k-15 + 1 = (3k-14)²

 1k·10k-15 + 1 = (10k-1)/9 × (10k + 5) + 1 = (102k - 10k + 5·10k - 5) / 9 + 1 = (102k + 4·10k + 4) / 9 = [(10k + 2) / 3]².

9k-1820k-181 = (9k1)².

 9k1² = (10(10k - 1) + 1)² = (10k+1 - 9)² = 102k+2 - 18·10k+1 + 81 = 102k+2 - (20 - 2)·10k+1 + 81 = 102k+2 - 2·10k+2 + 2·10k+1 + 81 = 10k+2(10k - 2) + 2·10k+1 + 81 = 10k+2(10(10k-1 - 1) + 8) + 2·10k+1 + 81 = 10k+3(10k-1 - 1) + 8·10k+2 + 2·10k+1 + 81 = 10k+3(10k-1 - 1) + 82·10k+1 + 81 = 9k-1820k-181.