Sum Square Factorials

Number Theory Level 3

There are sums of distinct factorials which give square numbers.

A solution is: $2^2 = 0!+1!+2!$ in the form $a^2=b!+c! \dots n!$

How many positive solutions are there for $a < 10^{12}$ (excluding the one shown above)?

Note :

The factorials need not be consecutive.
Factorials can only appear once in a solution but the value of the factorials may be the same. (eg. we can write $0!+1!+2!=2^2$ but not $1!+1!+2!=2^2$ .
Factorials in this question are only non-negative integers. (For eg. We can use $1!$ but not $1.5!$ or $-1.5!$ )
$0!$ is taken as $1!$ only if one of them is used in a solution. -You cannot substitute factorials as the sum of other factorials unless there is no other way. $eg, 2!=1!+0!$ so if $2!$ does not repeat then the substitute solution is not taken into account. (Because we strive for lesser terms to achieve accuracy and efficiency.)

Bonus : Are there infinite solutions for all real numbers. Prove it.

Try hard!

The answer is 13.

1 solution

Mohammad Farhat
Jun 14, 2019

D. Hoey listed sums $<10^{12}$ of distinct factorials which give square numbers

$0!+1!+2!=2^2$ $1!+2!+3!=3^2$ $1!+4!=5^2$ $1!+5!=11^2$ $4!+5!=12^2$ $1!+2!+3!+6!=27^2$ $1!+5!+6!=29^2$ $1!+7!=71!$ $4!+5!+7!=72^2$ $1!+2!+3!+7!+8!=213^2$ $1!+4!+5!+6!+7!+8!=215^2$ $1!+2!+3!+6!+9!=603^2$ $1!+4!+8!+9!=615^2$ $1!+2!+3!+6!+7!+8!+10!=1917^2$

Just for fun,

$1!+2!+3!+7!+8!+9!+10!+11!+12!+13!+14!+15!=1183893^2$

Sum Square Factorials

The answer is 13.

1 solution

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