+2 @geometry dash Let P(m,n) denote the assertion that
f(m+n)=f(m)+f(n)−2f(mn+m+n+1)+m2+n2.f(m+n) = f(m) + f(n) - 2f(mn + m + n + 1) + m^2 + n^2.f(m+n)=f(m)+f(n)−2f(mn+m+n+1)+m2+n2.
Start with P(0,0):
f(0+0)=f(0)+f(0)−2f(0+0+0+1)+02+02.f(0 + 0) = f(0) + f(0) - 2f(0 + 0 + 0 + 1) + 0^2 + 0^2.f(0+0)=f(0)+f(0)−2f(0+0+0+1)+02+02.
This simplifies to:
f(0)=2f(0)−2f(1).f(0) = 2f(0) - 2f(1).f(0)=2f(0)−2f(1).
Given that f(1)=0f(1) = 0f(1)=0, we have:
f(0)=2f(0)−0 ⟹ f(0)=2f(0).f(0) = 2f(0) - 0 \implies f(0) = 2f(0).f(0)=2f(0)−0⟹f(0)=2f(0).
This implies f(0)=0.
Next, we apply P(m,0):
f(m+0)=f(m)+f(0)−2f(m⋅0+m+0+1)+m2+02.f(m + 0) = f(m) + f(0) - 2f(m \cdot 0 + m + 0 + 1) + m^2 + 0^2.f(m+0)=f(m)+f(0)−2f(m⋅0+m+0+1)+m2+02.
This simplifies to:
f(m)=f(m)+0−2f(m+1)+m2.f(m) = f(m) + 0 - 2f(m + 1) + m^2.f(m)=f(m)+0−2f(m+1)+m2.
Rearranging gives:
0=−2f(m+1)+m2 ⟹ 2f(m+1)=m2 ⟹ f(m+1)=m22.0 = -2f(m + 1) + m^2 \implies 2f(m + 1) = m^2 \implies f(m + 1) = \frac{m^2}{2}.0=−2f(m+1)+m2⟹2f(m+1)=m2⟹f(m+1)=2m2.
To express f(m), we note that m+1m + 1m+1 can be replaced by nnn, so:
f(n)=(n−1)22.f(n) = \frac{(n-1)^2}{2}.f(n)=2(n−1)2.
Substituting n=m+1n = m + 1n=m+1:
f(m)=(m−1)22.f(m) = \frac{(m-1)^2}{2}.f(m)=2(m−1)2.
We can compute f(m)f(m)f(m) for any nonnegative integer mmm:
For m=0m = 0m=0:
f(0)=(−1)22=12.f(0) = \frac{(-1)^2}{2} = \frac{1}{2}.f(0)=2(−1)2=21.
For m=1m = 1m=1:
f(1)=0(given).f(1) = 0 \quad \text{(given)}.f(1)=0(given).
For m=2m = 2m=2:
f(2)=(2−1)22=12.f(2) = \frac{(2-1)^2}{2} = \frac{1}{2}.f(2)=2(2−1)2=21.
For m=3m = 3m=3:
f(3)=(3−1)22=42=2.f(3) = \frac{(3-1)^2}{2} = \frac{4}{2} = 2.f(3)=2(3−1)2=24=2.
For m=4m = 4m=4:
f(4)=(4−1)22=92.f(4) = \frac{(4-1)^2}{2} = \frac{9}{2}.f(4)=2(4−1)2=29.
We can generalize to obtain:
f(n)=(n−1)22.f(n) = \frac{(n-1)^2}{2}.f(n)=2(n−1)2.
Now we compute f(123)f(123)f(123):
f(123)=(123−1)22=12222=148842=7442.f(123) = \frac{(123 - 1)^2}{2} = \frac{122^2}{2} = \frac{14884}{2} = 7442.f(123)=2(123−1)2=21222=214884=7442.