 We have
f (n) ≔n^{4} +6n^{3} +11n^{2} +3n + 31 = (n^{2} +3n + 1)^{2}  3(n  10).
Therefore f (10) is a perfect square, and we now show there is no
other integer n such that f (n) is a perfect square.
We have
(n^{2} +3n + 2)^{2}  (n^{2} +3n + 1)^{2} = 2n^{2} + 6n + 3 and
(n^{2} +3n + 1)^{2}  (n^{2} +3n)^{2} = 2n^{2} + 6n + 1.
We have four cases to consider.
 (a)
 n > 10. Then we have
3(n  10)≥2n^{2} + 6n + 1, which is not
possible.
 (b)
 2 < n < 10. Then we have
3(10  n)≥2n^{2} + 6n + 3, which is not
possible.
 (c)
 n <  6. Then we have
3(10  n)≥2n^{2} + 6n + 3, which is not
possible.
 (d)

6≤n≤2. Then we can check individually that the 9 values
of n do not make f (n) a perfect square.
We conclude that f (n) is a perfect square only when n = 10.
 The folded 3dimensional region can be described as a
regular tetrahedron with four regular
tetrahedrons at each vertex cut off. The four smaller
tetrahedrons have side length 2cm., while the big tetrahedron
has sides of length 6cm. Recall that the volume of a regular
tetrahedron of side of length 1 is
√2/12 (or easy
calculation). Therefore the volume required in cm^{3} is
6^{3}√2/12  4·2^{3}√2/12 = 46√2/3.
 Let n = 2015. If one regards
a_{1},..., a_{n} as variables, the
determinant is skew symmetric (i.e. if we interchange a_{i} and
a_{j} where
i≠j, we obtain det A). We deduce that a_{i}  a_{j} divides det A for all
i≠j, hence
det A is divisible by ∏_{1≤i < j≤n}(a_{i}  a_{j}).
For
k∈ℕ, we prove by induction on k that if a number
is divisible
a_{1}...a_{k} and
∏_{1≤i < j≤k}(a_{i}  a_{j}), then it
is divisible by k!; the case k = 1 is immediate. So assume the
result for
i≤k. If one of the a_{i} is divisible by k + 1, then
the result is true for k + 1 by induction. On the other hand if none
of the a_{i} is divisible by k + 1, then at least one of the numbers
a_{i}  a_{j} is divisible by k + 1 and the induction step
is complete. The result follows.
 We first show the result is true if
0 < p≤1 for
p∈ℚ (positive number excludes 0, however the result is even
true here by taking the sum of a zero number of terms). Write p = a/b where
a, b∈ℕ. The result is obviously true if a = 1. We now prove the result by induction on a; we may assume
that a < b. Let
n≥2 be the unique positive integer such that
1/n≤p < 1/(n  1). Then we have
b≤an and
0≠an  a < b.
Set
q = (an  b)/bn. Since an  b < a, we may write q as a partial
sum S of the 1/m, and then we have
p = S + 1/n. Also the
integers m which appear in S must have m > n, because p < n  1.
This completes the induction step, and we have proven the result for
p≤1.
Let
p∈ℚ and let
s_{n} = ∑_{k=1}^{n}1/k. Since the
harmonic series is divergent, there exists a unique
m∈ℕ such that
s_{m} < p≤s_{m+1}. Then p  s_{m} < 1, so by
the previous paragraph is a partial sum S of the 1/n, and then we
have
p = S + 1/m. Also
S≤1/(m + 1) so none of the 1/n
appearing in S can be equal to 1/m, and the proof is complete.
 Let n be a positive integer. Then
∫_{0}^{n}∫_{1}^{π}1/(1 + (xy)^{2}) dxdy = ∫_{1}^{π}∫_{0}^{n}1/(1 + (xy)^{2}) dydx.
Therefore
∫_{0}^{n}(arctan(πx)  arctan(x))/x dx = ∫_{1}^{π}(arctan(ny))/y dy.
Set
u = arctan(ny) and dv = 1/y and use integration by parts to
obtain
∫_{1}^{π}arctan(ny)/y dy = arctan(nπ)lnπ  ∫_{1}^{π}(n ln y)/(1 + n^{2}y^{2}) dy.
On the other hand,
0≤(n ln y)/(1 + n^{2})≤(n lnπ)/(1 + n^{2}) for all
y∈[1,π]. Therefore
lim_{n→∞}∫_{1}^{π}(arctan(ny))/y dy = (πlnπ)/2
and we deduce that
∫_{0}^{∞}(arctan(πx)  arctan(x))/x dx = lim_{n→∞}∫_{0}^{n}(arctan(πx)  arctan(x))/x dx = (πlnπ)/2.
 If
(x, y)∈S ≔∑_{i=1}^{n}ℤ(a_{i}, b_{i}), then there
exist
k_{i}∈ℤ such that
(x, y) = ∑_{i=1}^{n}k_{i}(a_{i}, b_{i}). We choose the k_{i} such that
d ≔∑_{i=1}^{n} k_{i} is minimal and then define
d (x, y) = d. On the other hand if
(x, y)∉S, then define
d (x, y) = + ∞ (thus
d (x, y) = ∞ if and only if
(x, y)∉S). Now choose a
positive integer m such that
m≥n/ϵ and define
f (x, y) = 1  d (x, y)/m
if
d (x, y)≤m and
0 if
d (x, y) > m.
If
(x, y)∉S, then
(x + a_{i}, y + b_{i})∉S for all i and
therefore
d (x, y) = d (x + a_{i}, y + b_{i}) = 0 and hence
f (x, y) = 0 if
d (x, y) = + ∞. On the other hand if
(x, y)∈S, then
 d (x, y)  d (x + a_{i}, y + b_{i})≤1 for all i.
It then follows that
f (x, y) = 0 if
d (x, y) > m, hence
f (x, y)≠ 0 for only finitely many (x, y),
and furthermore
 f (x, y)  f (x + a_{i}, y + b_{i}) = 0 or 1/m for all
i. Thus f (x, y) satisfies the required condition, so the answer
is ``yes".
 Note that the hypotheses show that there exists a positive integer
a such that
a⟨u, v⟩∈ℤ for all
u, v∈S. Therefore there exists a positive integer b such that
b u^{2} = b⟨u, u⟩ is a positive
integer for all
0≠u∈S, so we may choose
0≠s∈S
such that  s is minimal.
First suppose that the x_{i} are all contained in
ℝs
(i.e. the points of S are collinear).
Then the same is true of S and we claim that
S = ℤs.
If
u∈S, then
u = cs for some
c∈ℝ. Also
a s≤ u < (a + 1) s for some nonnegative integer a,
hence
 as≤ cs < (a + 1)s. We deduce that
(a  c)s <  s and since
(a  c)s∈S, we conclude that
(a  c)s = 0.
Therefore u = as and the claim is established. Now we place
disks of radius
R ≔3 s/4 and center
(2n + 1/2)s for all
n∈ℤ and the result is proven in this case.
Now suppose that not all the x_{i} are not contained
ℝs.
Then we may choose
t∈S\ℝs with
 t minimal. We claim
that
S = T ≔{ms + nt  m, n∈ℤ}. If this is not
the case, we may choose
u∈S\T.
Note that
ℝs + ℝt = ℝ^{2}, so we may write
u = ps + qt for some
p, q∈ℝ and then there exist
a, b∈ℤ such
that
a≤p < a + 1 and
b≤q < b + 1, so u is inside the
parallelogram with vertices (as, bt),
(as + s, bt), (as, bt + t),
(as + s, bt + t). Since
 s≤ t we see that u is
distance at most  t from one of these vertices. Furthermore
 u  v≥ t for all
u≠v∈S, so we must have
u∈S.
Now we can place disks with radius
R ≔ s/2 and centers at
((2m + 1/2)s, nt) for
m, n∈ℤ. Clearly every disk
contains at least two points of S, namely (2ms, nt) and
(2ms + 1, nt) for the disk centered at
((2m + 1/2)s, nt), and these
disks accounts for all the points in S. We only need to show that
any two distinct disks intersect in at most one point, and thus we
need to show that two different centers are distance at least  s
apart. So consider two different centers, say at
((2m + 1/2)s, nt)
and
((2m' + 1/2)s, nt'). Then the distance between theses two
centers is the same at the distance between (2ms, nt) and
(2m's, n't), which is at least  s by minimality of  s.
This completes the proof. (This actually proves the stronger
statement that every point of S lies in exactly one disk, which is
how the problem was meant to be stated; the argument can be
significantly shortened for the actual problem.)