Uniquely Awesome

The given equation can be put in the form :

n (n + 1) (n^2 + 1) =(n - 1)! (n + 2)(n + 1)(n)

(n - 1)! = (n^2 + 1) / (n + 2) = (n - 2) + 5 / (n + 2)

Then

5 / (n + 2) must be an integer

So

n + 2 = 1 ............... i e n = -1 (refused)

Or

n + 2 = 5 .............. i e n = 3

Start by factoring the given expression,after you are done you would have got: $[n(1+n^2)(1+n)]=(n+2)!\\ \Longrightarrow (n^2+1)=(n+2)(n-1)!(\text{cancelling 'n' and '(n+1)'from L.H.S and R.H.S)}\\\Longrightarrow \dfrac{n^2+1}{n-1}=(n+2)(n-2)!=\text{an integer}\\ \Longrightarrow (n-1)|(n^2+1)\\ \Longrightarrow (n-1)|n(n-1)+n+1\\ \Longrightarrow (n-1)|(n+1)\\ \Longrightarrow (n-1)|(n-1)+2\\ \Longrightarrow (n-1)|2$ After checking all the possible cases,only $3$ satisfies the given conditions.

$n + n^2 + n^3 + n^4 = (n + 2)!$

$n^2 + n^3 + n^4 + n^5 = n(n + 2)!$

$(n + 2)! - n(n + 2)! = n - n^5$

$(1 - n)(n + 2)! = n(1 - n^4)$

$(n + 2)! = \frac{n(1 + n^2)(1 - n^2)}{1 - n}$

$(n + 2)! = \frac{n(1 + n^2)(1 + n)(1 - n)}{1 - n}$

$(n + 2)! = n(n + 1)(n^2 + 1)$

$(n + 2)(n + 1)n(n - 1)! = n(n + 1)(n^2 + 1)$

$(n + 2)(n - 1)! = n^2 + 1$

$(n - 1)! = \frac{n^2 + 1}{n + 2}$

$(n - 1)! = \frac{(n^2 + 1) + (4 - 4)}{n + 2}$

$(n - 1)! = \frac{(n^2 - 4) + (4 + 1)}{n + 2}$

$(n - 1)! = \frac{(n - 2)(n + 2) + 5}{n + 2}$

$(n - 1)! = (n - 2) + \frac{5}{n + 2}$

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$n$ is an integer $\therefore (n + 2)$ is also an integer. $\frac{5}{n + 2}$ is also an integer.

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$n + 2 = -5,~ -1,~ 1,~ 5 \rightarrow n = -7,~ -3,~ -1,~ 3$

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The factorial function is defined for all non-negative integers:

$(n - 1) \geq 0 \therefore n \geq 1$

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$n = 3$

$3 + 3^2 + 3^3 + 3^4 = (3 + 2)!$

$120 = 120$

$\large \boxed{n = 3}$

Hence, the result has to be "n=3"

Okay, so first let's write the right hand side in a more suitable way:

$n + n^{2} + n^{3} + n^{4} = (n + n^{2}) + n^{2}*(n + n^{2}) = (n + n^{2})*(1 + n^{2}) = n*(n + 1)*(n^2 + 1)$

From this, divide both sides by the product $n*(n+1)*(n+2)$

We then get:

$\frac{n^{2} + 1}{n + 2} = (n - 1)!$

The left hand side is an integer, and so must be the right hand side. Let's once more re-write the left hand side in a more suitable fashion:

$\frac{n^{2} + 1}{n + 2} = \frac{n^{2} + 2n - 2n - 4 + 4 + 1}{n + 2} = \frac{n*(n+2) - 2(n + 2) + 5}{n + 2} \rightarrow (n -2) + \frac{5}{n + 2}$

Rearranging the equation, we get:

$\frac{5}{n + 2} = (n-1)! - (n - 2)$

The right hand side must still be an integer, and so must be the left hand side. Now, notice that the factorial function is only defined for non-negative integers; thus, from the original equation, we must have as a first condition: $n + 2 \geq 0 \rightarrow n \geq -2$ .

Our second condition lies within the fact that $n + 2$ must be a divisor of $5$ - which we got from rearranging the terms - which means that $n + 2$ has to be an element of the set $\pm 1, \pm 5$ . From this, we derive that n belongs to the set ${-7, -3, -1, 3}$ .

Given both constraints, we have that $n$ is either $-1$ or $3$ . Let's test both possibilities.

For $n = -1$ , we have: $(-1) + (-1)^{2} + (-1)^{3} + (-1)^{4} = ((-1) + 2)! \rightarrow -1 + 1 - 1 + 1 = (1)! \rightarrow 0 = 1$ , which is a false claim and therefore $n = -1$ is not a valid solution.

For $n = 3$ , we have: $3 + 3^{2} + 3^{3} + 3^{4} = (3 + 2)! \rightarrow 3 + 9 + 27 + 81 = 5! \rightarrow 120 = 120$ which is true, and therefore not only $n = 3$ is a solution, but also a unique one.

The value of n is 3

3+9+27+81=(5)!

120=120

You assume that it is an odd integer and a prime because prime integers are odd numbers (as in different from the rest) so you take the first "odd" integer which ends up being 3 since 1 is not prime. It isn't brute force, it's merely being uniquely awesome.

n + n^2 + n^3 + n^4 = (n + 2)!

n(1 + n + n^2 + n^3) = (n + 2)!

Factorise LHS and divide through by n;

1 + n + (n^2)(1 + n) = (n + 2)!/n

Further factorise LHS;

1 + n + (n^2)(1 + n) = (n^2 + 1)(n + 1)

For RHS; (n + 2)! = (n + 2)(n +1)(n)(n - 1)!

So; (n + 2)! / n = (n + 2)(n + 1)(n - 1)!

Then LHS = RHS is;

(n^2 + 1)(n + 1) = (n + 2)(n + 1)(n - 1)!

Cancel n + 1 off both sides gives;

(n^2 + 1) = (n + 2)(n - 1)!

Note; n^2 + 1 = n^2 - 4 +5

And; n^2 - 4 = (n + 2)(n - 2)

So (n^2 + 1) = (n + 2)(n - 1)! becomes;

(n + 2)(n - 2) + 5 = (n + 2)(n - 1)!

[(n + 2)(n - 2) + 5] / (n + 2) = (n - 1)!

n - 2 + [5 / (n + 2)] = (n - 1)!

Now; 5 / (n + 2) must be an integer

Let 5 / (n + 2) = k, k an integer

n = (5 / k) - 2

If n & k are integers, k must be; -5, 1, or 5 only.

Then; n = -3, 3, or - 1

Consider LHS; (n - 1)! ... Negative numbers give indefinite factorials; therefore n = -3 or -1 cannot be correct.

n = 3 is the correct solution.

One can prove that for n>=4 there is no solution as RHS>LHS (not a hard problem) There is also no solution for n <=-3 as negative factorials are not defined. Testing -2,-1,0,1,2,3 gives 3 as the solution .

I just typed 3 because it seemed like the coolest answer.

Simply start bashing and you'll finish at $\boxed{3}$