Show the Lagrange Four Square Theorem for
For any natural number (p), we write as
p = a2 + b2 + c2 + d2
Floor(√) = Floor()
Floor() = 0
This is called max_a
Find the first value of a such that
a2 ≥ n/4
Start with min_a = 0 and increase by 1
Continue until we reach or breach n/4 → /4 = 0
When min_a = 0, then it is a2 = 1 ≥ 0, so min_a = 0
(0, 0)
Find max_b which is Floor(√n - a2)
max_b = Floor(√ - 02)
max_b = Floor(√ - 0)
max_b = Floor(√0)
max_b = Floor(0)
max_b = 0
Find b such that b2 ≥ (n - a2)/3
Call it min_b
Start with min_b = 0 and increase by 1
Go until (n - a2)/3 → ( - 02)/3 = 0
When min_b = 0, then it is b2 = 1 ≥ 0, so min_b = 0
(0, 0)
Determine max_c =Floor(√n - a2 - b2)
max_c = Floor(√ - 02 - 02)
max_c = Floor(√ - 0 - 0)
max_c = Floor(√0)
max_c = Floor(0)
max_c = 0
Call it min_b
Start with min_c = 0 and increase by 1
Go until (n - a2 - b2 )/2 → ( - 02 - 02)/2 = 0
When min_c = 0, then it is c2 = 1 ≥ 0, so min_c = 0
See if d is an integer solution which is √n - a2 - b2
max_d = √ - 02 - 02 - 02
max_d = √ - 0 - 0 - 0
max_d = √0
max_d = 0
Since max_d = 0, then (a, b, c, d) = (0, 0, 0, 0) is an integer solution proven below
02 + 02 + 02 + 02 → 0 + 0 + 0 + 0 =
(a, b, c, d) = (0, 0, 0, 0)
(a, b, c, d) = (0, 0, 0, 0)
Free Lagrange Four Square Theorem (Bachet Conjecture) Calculator - Builds the Lagrange Theorem Notation (Bachet Conjecture) for any natural number using the Sum of four squares.
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