Firstly being in the Number Theory section, I do not believe this is a sufficient solution a) this is coding b) you did not show there are no other solutions.

A favourite technique to prove some number is not a square (for most integers) is to bound it between two squares. Write $m=1+n+n^2+n^3+n^4$ (assuming $n$ is an integer of course).

For the lower bound, we know we want some square $(n^2+k)^2<m$ . Certainly $\frac{n}{2}$ would appear in $k$ as we want a strong bound that includes $n^3$ in it. So if we try $(n^2+\frac{n}{2})^2=n^4+n^3+\frac{n^2}{4}<m$ , we now have a reasonable lower bound.

Since we do not know if $n$ is odd or even, the only choice is to make the upper bound $(n^2+\frac{n+1}{2})^2>m$ and hope that it works. Expanding gives $n^4+n^3+n^2+\frac{n^2+2n+1}{4}$ . But to make this larger than $m$ we now just need $n^2+2n+1>4(n+1) \iff n^2-2n-3=(n-3)(n+1)>0 \iff n<-1$ or $n>3$ . So lastly test $n=-1,0,1,2,3$ which gives values $1, 1, 5, 31, 121$ .

The reason this works is because for those specified $n$ , since there is no integer between $\frac{n}{2}$ and $\frac{n+1}{2}$ , therefore $m$ cannot be an integer square!

So we know for certain $1$ and $121$ are the only squares, and the answer is the sum of idigits in $11$ , i.e. $2$ .

For me I let the perfect square be $m^2$ . So now $m^2$ - 1 = n+ $n^2$ + $n^3$ + $n^4$ or (m+1)(m-1) = n(1+n+ $n^2$ + $n^3$ ) = n(1+n)(1+ $n^2$ ).

Now it is pretty obvious that n(1+n)(1+ $n^2$ ) is even for all values of n right? But not all even values would give an integer value for m as if (m+1)(m-1) is divisible by 2 but not 4, either m+1 or m-1 is even and the other odd. hence n(1+n)(1+ $n^2$ ) should be divisible by 4.

But one thing to note is that n(1+n)(1+ $n^2$ ) cannot be exactly 4 as m+1=2 and m-1 = 2 which is absurd. So n cannot be 1 here. Also, n=2 would not produce a multiple of 4. So smallest n=3 so that we have (m+1)(m-1) = 3(3+1)( $3^2$ +1) = 3(4)(10) = 120.

Now we have m+1 = 12 and m-1 = 10 so that m=11. Then 1+1 = $\boxed{2}$ .

Python 2.7:

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def f(i):
    return 1 + i + i**2 + i**3 + i**4
def in_form(square):
    i = 0
    while f(i) <= square:
        if f(i) == square:
            return True
        i += 1
    return False
n = 2
while not in_form(n*n):
    n += 1
print sum([int(i) for i in str(n)])

121 squrt(121)=11 1+1=2

let x be that number, then LHS is (n^5-1)/(n-1), the sum of GP , then LHS=x^2, put n=3, you'll get x to be 11. So the sum of digits is 2. Of course this solution is based on a good guess. But more rigorous analysis of this problem can also be done, which in fact some of the people have done it in answer section.

Lazy

1 + 1 = 2