Question

Show that 1! + 2! + 3! + ... + n! Cannot be a perfect square for any n € N, n ≥ 4

Answer 1

'n' is a natural number which is greater than or equal to 4.

Consider n = 4.

1! + 2! + 3! + 4! = 1 + 2 + 6 + 24 = 33.

33 is not a perfect square. Thus proved it when n = 4.

We are familiar with the concept that natural numbers ending in 2, 3, 7 and 8 are not perfect squares.

Here, when taking n = 4, the sum got is 33, which ends in 3. Thus it wouldn't be a perfect square.

As 33 ends in 3, let 1! + 2! + 3! + 4! be 10k + 3, where k = 3.

Consider n = 5.

1! + 2! + 3! + 4! + 5!

⇒ 10k + 3 + 120

⇒ 10k + 3 + 12 × 10

⇒ 10(k + 12) + 3     →   (1)

Here, it also ends in 3, thus it wouldn't be a perfect square even when n = 5.

Consider n = 6.

1! + 2! + 3! + 4! + 5! + 6!

⇒ 10(k + 12) + 3 + 720          [From (1)]

⇒ 10(k + 12) + 3 + 72 × 10

⇒ 10(k + 12 + 72) + 3

⇒ 10(k + 84) + 3.

Here, also. Thus neither would be it when n = 6.

We found that 1! + 2! + 3! +......+ n! is not a perfect square where n = 4. Taking n greater than 4, n! contains 5 to be multiplied and any even number, thus n! ends in 0. Hence, as the sum where n = 4 ends in 3, then so will be the sums where n > 4.

Hence proved that 1! + 2! + 3! +......+ n! can't be a perfect square where n ∈ N and n ≥ 4.