Look at my solution here

We have $ n^2 - 10n - 22 \le 9^{log (n - 1)}$ Because n has at most log (n-1) digits, which are at most 9. Take the log of both sides: $ \frac {log(n^2 - 10n - 22)}{log (n - 1)} \le log 9$ Log 9 is less than 1, so we have $ n^2 - 10n - 22 \le n - 1 \implies n\le 12$ However, we must have $ n^2 - 10n - 22 \ge 0 \implies n \ge 12$, so $ n = 12$ is the only possibility and it checks.

let n=a0+a1*10+..+ar*10^r then n^2-10n-22=a0*a1*..*ar<10^(r+1) (and ai>0 for every i belonging to the set {0,1,..,r} as the equation x^2-10x-22=0 has no integral solution) since n^2-10n-22>=10^(2r) when r>1 2r<=r+1 or r<=1 as r cannot be less than 1 it has to be 1. then let the no. be 10a+b. then ab=(10a+b)^2-10(10a+b)-22 or, -19ab=100a*(a-1)-(10-b)*b-22 so a has to be 1, otherwise the RHS is positive while the LHS is negative. hence -19b=b^2-10b-22 or, (b-2)(b+11)=0 so b=2 pardon me for the way i've written the solution.i don't know where to get mathematical symbols and also i've never had any higher mathematics books, which hinders me from presenting an elegant solution.

It can be shown by induction on the number of digits of n that p(n) ≤ n for all n ∈ N. It follows that n^2 −10n−22 ≤ n which implies n ≤ 12. Since 0 < n^2 −10n−22 = (n−12)(n+2)+2, one easily obtains n ≥ 12. Now one can directly check that $n = 12$ is indeed a solution, and thus the only one.

dgreenb801 wrote: We have $ n^2 - 10n - 22 \le 9^{log (n - 1)}$ Because n has at most log (n-1) digits, which are at most 9. Take the log of both sides: $ \frac {log(n^2 - 10n - 22)}{log (n - 1)} \le log 9$ Log 9 is less than 1, so we have $ n^2 - 10n - 22 \le n - 1 \implies n\le 12$ However, we must have $ n^2 - 10n - 22 \ge 0 \implies n \ge 12$, so $ n = 12$ is the only possibility and it checks. But, how you are saying this " n has at most $\log{(n-1)} $ digits" ?

Another solution would be seeing that \(n^2-10n-22\) can never be a multiple of \(3,4,5\) nor \(7\) for integer \(n\), so the only possibilities for the digits of \(n\) are \(1,2\). From the expression, and our conclusion that the product of the integers can't be a multiple of \(4\), we see that if \(n\) is even, then it a sequence of \(1's\) followed by a \(2\). Else, it is a sequence of \(1's\). Hence the expression equals \(1\) or \(2\). Using the quadratic formula, we get the only solution is \(\boxed{12}\).

I will show that 12 is the only natural number that satisfies these requirements. We can easily see that $x=12$ satisfies these requirements. Let $f(x)$ be the quadratic $x^2 - 10x -22$. First we will find the roots of $f(x)$. We see that they are $5 + \sqrt{47}$ and $5 - \sqrt{47}$. These rounded to the nearest tenth are 11.9 and -1.9, respectively. Since the $x^2$ coefficient of $f(x)$ is positive, the $x$ values in between the roots will output negative $f(x)$ values. Since the product of the digits must be non-negative, $x>11.9$. Claim: If $x>12$, $f(x)>x$ Proof: First, note that $f(13)>13$. Now, note that as $x$ increases by 1, $10x$ increases by 10 and $x^2$ increases by 2x+1, so $f(x)$ increases by $2x-9$. So $f(x)$ increases by more than 1 as long as $x>5$. Therefore, our claim is true. Claim: If $x>9$, the product of the digits of $x$ is less than $x$. Proof: For any number with the decimal representation $a_{i-1}\cdots a_0$, the product of the digits is bounded by $a_{i-1} 9^{i-1}$, whereas the number is bounded from below by $a_{i-1} 10^{i-1}$. Therefore, if $i>0$, which is when $x>9$, our claim is true. So if $x>13$, $f(x)>x>$ the product of the digits of $x$, which means that $f(x)$ ≠ $x$, so we are done. (The only integer $x$ such that $11.9<x<13$ is $\boxed{12}$, which works)