Proving that $\sum\limits_{i=1}^k i! \ne n^2$ for any $n$

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How to prove that the number 1!+2!+3!+…+n! is never square?

Show that $\displaystyle\sum\limits_{i=1}^k i!$ is never a perfect square for $k\ge4$

I could prove $k!$ is never a perfect square using Bertrand's Postulate. But this one seems to be an uphill task.

score 43 · Answer 1 · edited Mar 08 '12 at 13:46

43

Your summation looks like $$1!+2!+3!+4!+\cdots+k!=33+5!+\cdots+k!$$Now note that $i!$ is a multiple of $10$ whenever, $i\geq 5$, hence the last digit in your summation is going to be a $3$. By inspection modulo $10$ (I want to look at the decimal place) $$0^2=0$$$$1^2=1$$$$2^2=4$$$$3^2=9$$$$4^2=6$$$$5^2=5$$$$6^2=6$$$$7^2=9$$$$8^2=4$$$$9^2=1$$Hence, it is impossible to have a $3$ in the far most right digit of a square, which your sum does.

edited Mar 08 '12 at 13:46

Michael Hardy

1

answered Mar 08 '12 at 08:43

Daniel Montealegre

6,783

You could have made a shorter list of squares by writing $(\pm 1)^2=1$, $(\pm 2)^2=4$ etc. – Ewan Delanoy Mar 08 '12 at 17:25
Nice—I just made 31 up-votes. :) – Nick Strehlke Mar 08 '12 at 20:45
Very concise! =) – Elchanan Solomon Mar 14 '12 at 05:25
2

$33$ is quite a lot of upvotes for a closed question -- and it also happens to be the sum of the first four factorials that you use in the proof :-) – joriki Mar 20 '12 at 08:28

draks ... · Answer 2 · 2016-07-12T05:15:27.507

12

Your last digit at $k=4\;:\; 1!+2!+3!+4!=33$ is $3$ and from $5!$ on you add numbers that end with $0$. So it can never be a square number, since the last digits of the square of any number ends with $(0^2,(\pm 1)^2,(\pm 2)^2,(\pm 3)^2,(\pm 4)^2,(\pm 5)^2)\mod 10=(0,1,4,9,6,5)$.

edited Jul 12 '16 at 05:15

answered Mar 08 '12 at 08:49

draks ...

18,449

It is further interesting to note, that $3$ and $11$ are always a factor of the sum ($k\geq 3$ in the case $3$ and $k\geq10$ in the case of $11$), due to the fact that $$ 1!+2!+3!=9 \text{ and } \sum_{k=1}^{10}k!=3^2×11×40787. $$ Every futher addend also shares a $3$ resp. $11$, e.g. $$ \begin{eqnarray} 1!+2!+3!+4!+\cdots &=& 3\times (1+2+\frac{4!}{3}+\cdots)\ (\sum_{k=1}^{10}k!)+11!+ \cdots&=& 11\times(3^2\times40787+\frac{11!}{11}+\cdots) \end{eqnarray} $$ If theses are the only values, is asked here. – draks ... Mar 16 '12 at 14:12

score 3 · Answer 3 · answered Mar 08 '12 at 14:45

3

Hint $\rm\ mod\ 5\!:\ \ \mathbb Z^2 \equiv \{0,\: \pm 1,\:\pm2\}^2 \equiv \{0,\: \pm 1\} \not\ni -2 \equiv 1!+2!+3!+4! + 5 N$

answered Mar 08 '12 at 14:45

Bill Dubuque

272,048

Eric Naslund · Answer 4 · 2013-01-17T23:58:03.480

1

The following is a brute force approach. We may calculate the sum at $n=21$, and find $$\sum_{k=1}^{21}k!=3^2\cdot 11\cdot 877\cdot3203\cdot41051\cdot4699727.$$

Since $11^2$ divides $k!$ for all $k\geq 22$, this implies that $11$ divides the sum exactly once for all $k\geq 21$, and hence the sum is not a square. For $n\leq 21$, we may check by computer and notice none are squares except $n=3$

edited Jan 17 '13 at 23:58

answered Mar 08 '12 at 08:45

Eric Naslund

72,099

Proving that $\sum\limits_{i=1}^k i! \ne n^2$ for any $n$

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