Finding $a$ and $b$ When $a^2+b^2=n!$ and $a,b,n \in N$

[Albert1]

$a,b,n \in N$ ,$a\leq b \,\, and \,\, n<14$
$if \,\ a^2+b^2=n!$
$find :\,\, a,b$

 

[HallsofIvy]

Spoiler

Since n< 14, it is not too difficult to go through the possible values:
n= 1, n!= 1. There are no positive integers, a and b, such that $$a^2+ b^2= 1$$.

n= 2, n!= 2. $$1^2+ 1^2= 2$$ so a= b= 1 is a solution.

n= 3, n!= 6. $$1^2+ 5= 6$$ but 5 is not a square. $$2^2+ 2= 6$$ but 2 is not a square. There are no positive integers, a and b such that $$a^2+ b^2= 6$$.

n= 4, n!= 24. $$1^2+ 23= 24$$ but 23 is not a square. $$2^2+ 20= 24$$ but 20 is not a square. $$3^2+ 15= 24$$ but 15 is not a square. $$4^2+ 8= 24$$ but 8 is not a square. There are no positive integers, a and b such that $$a^2+ b^2= 24$$.

n= 4, n!= 120. $$1^2+ 119= 120$$ but 119 is not a square. $$2^2+ 116= 120$$ but 116 is not a square. $$3^2+ 111= 120$$ but 111 is not a square. $$4^2+ 104= 120$$ but 104 is not a square. $$5^2+ 95= 120$$ but 95 is not a square. $$6^2+ 84= 120$$ but 84 is not a square. $$7^2+ 71= 120$$ but 71 is not a square. $$8^2+ 56= 120$$ but 56 is not a square. There are no positive integers, a and b, such that $$a^2+ b^2= 4!$$

Etc. tedious but doable.

It might be simpler to think in terms of "Pythagorean triples": 2, 3, 5, or 5, 12, 13, and multiples of that.

 

[Opalg]

$a,b,n \in N$ ,$a\leq b \,\, and \,\, n<14$
$if \,\ a^2+b^2=n!$
$find :\,\, a,b$

[sp]A condition for an integer to be the sum of two squares is that each prime factor of the form $4k+3$ should occur to an even power. After $2 = 1^2 + 1^2$, the next few factorials have a factor $3$ (occurring just once), which prevents them being the sum of two squares. The first factorial to have a repeated factor $3$ is $6! = 720 = 2^4\cdot3^2\cdot5$, so that is a sum of two squares. And in fact $720 = 144 + 576 = 12^2 + 24^2$. After that, the factor $7$ comes in, and occurs just once in all the factorials up to $13!$. So none of these will be a sum of two squares.

Will there ever be any more factorials that are the sum of two squares? With primes such as $11$, $19$, $23\ldots$, all of the form $4k+3$, coming into play, it will be a very long time until a factorial occurs in which each of them occurs an even number of times.[/sp]