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# Squaresearch

Pete Inglesby from Wolverhampton; Patrick Snow, age 16, Dame Alice Owens School, Potters Bar; Yatir Halevi, age 18, Maccabim-Reut High-School, Israel; Andrei Lazanu, age 12, School 205, Bucharest, Romania all solved this problem. Here is Pete's solution:

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Pete Inglesby from Wolverhampton; Patrick Snow, age 16, Dame Alice Owens School, Potters Bar; Yatir Halevi, age 18, Maccabim-Reut High-School, Israel; Andrei Lazanu, age 12, School 205, Bucharest, Romania all solved this problem. Here is Pete's solution:

Consider numbers of the form

$$ u(n) = 1! + 2! + 3! +...+ n!. $$

How many such numbers are perfect squares?

In fact only when $n=1$ or $n=3$ does $u(n)$ give a perfect
square ($1$ and $9$ respectively).

The solution lies in the fact that all factorials above $5!$
are multiples of $10$. This is because they all have $2$ and $5$ as
prime factors. For $n> 5$

$$\eqalign { n!&=1 \times 2\times 3\times 4\times 5\times
\cdots \times n \cr &= 2\times 5(1\times 3\times 4\times \cdots
\times n) \cr &= 10(1\times 3\times 4\times \cdots n).}$$

When adding a multiple of $10$, to another integer, $k$, the
last digit of $k$ will remain the same. Suppose $k = 10j + l$ and
$m =10n$, with $l$ and $n$ integers, then $k + m = 10(j + n) +
l$.

So when adding all factorials equal to or higher than $5!$
then you are adding a multiple of ten, so the last digit of the sum
will be the same. This is important because the last digit of
$u(4)$ is a $3$. We have $u(4) = 1! + 2! + 3! + 4! = 1 + 2 + 6 + 26
= 33$. Therefore for any $u(n)$ with $n> 5$ the last digit will
be $3$. For any integer $p = (10a + b)$ with $a$ and $b$ integers
and $b< 10$, $p^2 = (100a^2 + 20ab + b^2)$. Therefore the last
digit is of $p^2$ is only dependent on $b^2$.

b | b^{2} |
Last digit of b^{2} |

0 | 0 | 0 |

1 | 1 | 1 |

2 | 4 | 4 |

3 | 9 | 9 |

4 | 16 | 6 |

5 | 25 | 5 |

6 | 36 | 6 |

7 | 49 | 9 |

8 | 64 | 4 |

9 | 81 | 1 |

This table shows that the squares of all integers between $0$ and $9$, and therefore all integers, end in either $1$, $4$, $5$, $6$ or $9$.

No perfect squares end in $3$. Therefore, $u(n)$ with $n > 4$ can never be a perfect square as it has been shown to always end in $3$.

All that remains is to find $u(n)$ for all integers of $4$ or less:

$u(1) = 1! = 1$;

$u(2) = 1! + 2! = 3$;

$u(3) = 1! + 2! + 3! = 9$;

$u(4) = 1! + 2! + 3! + 4! = 33.$

Only $u(1)$ and $u(3)$ are perfect squares, and so they are the only sums of factorials to be perfect squares.