How many natural numbers $n$ are there such that $n!+10$ is a perfect square

Question

How many natural numbers $n$ are there such that $n!+10$ is a perfect square?

The explanation given here is not very intuitive and I can't understand how to answer this question.

Answer 1

If you factor a perfect square into primes, every prime has an even exponent. The argument made says that for all $n \gt 5,$ there are at least two factors of $2$ in $n!$. In fact, this is true for all $n \ge 4$, but they covered $4$ and $5$ in the casework. Once $n!$ is a multiple of $4$, $n!+10\equiv 2 \pmod 4$ has an odd exponent of $2$, so it cannot be a perfect square.

Answer 2

Note that the perfect square numbers have the form $4k$ or $4k+1$ ($n=2m\to n^2=4m^2$ or $n=2m+1\to n^2=4(m^2+m)+1$).

If $n>3$, $n!$ is multiple of 4 and $n!+10\equiv 2 \mod 4$, an absurd.

Checking $n=1,2,3$, the only solution is $n=3$

Answer 3

The solution is saying:

$n! + 10= 2*3*4*5*6*......*n + 10$

$=10 (3*4*6*......*n +1) $

So $3*4*6*.....*n +1$ is odd so not divisible by 2..

$10$ is divisible by 1 power of two but not any higher power. So $n! +10$ is divisible by 2 but not 4. So it isn't a perfect square.

And yes that answer was awfully written.

The exception is if $n < 5$. Then $n! + 10 = 2*3*... + 10 = 2 (3*... +5) $ if n= 4 it is not prime as $3*4 +5$ is odd and we have $n! +10$ divisible by 2 but not 4. If $n=3$ we have $n! +10= 2*3 +10 =2(3+5)=2*8=16$ is square. $2! + 10=12$ and $1!+10=11$

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Oh, and for what it is worth, the other two answers are, in my opinion, more succinct and elegant than mine. But I wanted to specifically explain the reproduced answer sheet directly.

I had to read it three times before I could understand what "than exponant of 2 is one so it is not a perfact square" was supposed to mean. And I have no idea why it wrote that "1" making the expression $\frac{\text{[3 x 4 x 6 ....xn +1]}}{\text{1}}$ which is almost, but not quite, incomprehensible (Where did the "xn" come from ?).