smc

First, we wish to factor $P$ into a more manageable form. From the beginning of $P$, we notice $x^4+6x^3$, which gives us the idea to use $(x^2+3x{)}^2=x^4+6x^3+9x^2$. This gives us $$P=(x^2+3x{)}^2+2x^2+3x+31$$ This is not useful, but it gives us a place to start from. We can then try $(x^2+3x+1)=x^4+6x^3+11x^2+6x+1$. $$P=(x^2+3x+1{)}^2-3(x-10)$$ This is much more useful, as it moves all non-linear terms inside of a squared expression. We can then say $P=(x^2+3x+1{)}^2-3(x-10)=a^2$, where $a^2$ is the square of an integer mentioned on the problem. Right from here, we can set $x=10$, which cancels out the $3(x-10)$, giving $(100+30+1{)}^2=a^2$. This gives us one solution, $x=10$, $a=131$. We can then rearrange the expression, giving us $(x^2+3x+1{)}^2-a^2=3(x-10)$. Factoring using difference of squares, we obtain $$(x^2+3x+1+|a|)(x^2+3x+1-|a|)=3(x-10)$$ We can then state that when $x$ is greater than $3$ and less than $-9$, $x^2+3x+1$ will be greater than $|3x-10|$. This is obtained by setting $x^2+3x+1>|3x-10|$ and then solving the inequality. We can then conclude that $x^2+3x+1+|a|>|3x-10|$. Next, we claim that $x^2+3x+1-|a|\ge 1$ or $x^2+3x+1-|a|\le -1$ when $x\ne 10$. We can prove this by first noting that since $x$ and $a$ are integers, $x^2+3x+1-|a|$ is an integer. Next, we shall assume that $x^2+3x+1-|a|=0$. Solving this and plugging back into the original equation, we obtain $(2|a|)(0)=3(x-10)$. Solving we obtain $x=10$, which is a contraction to $x\ne 10$. Therefore, $x^2+3x+1-|a|\ne 0$ and $x^2+3x+1-|a|\ge 1$ or $x^2+3x+1-|a|\le -1$. Finally, we can go back to the equation $$(x^2+3x+1+|a|)(x^2+3x+1-|a|)=3(x-10)$$ We note that since $(x^2+3x+1+|a|)$ is larger than $3(x-10)$, in order for there to be solutions, $(x^2+3x+1-|a|)$ must be in the range $(-1,1)$. However, this contradicts what was proven earlier, so when $x\ne 10$ and $x<-9$ or $x>3$, there are no solutions for $x$. Now, all that remains to be checked are values of $x$ between $-9$ and $3$. Using brute force and checking each value individually, we can assert that there are no such solutions for $x$, leaving us with only $1$ solution, $x=10$. Therefore, the answer is $\boxed{1}$