Let N = 299...98200...029. There are n 9's and n 0's. Prove that N can be written as the sum of the squares of 3 consecutive natural numbers.
If x is the central of 3 consecutive natural numbers, the sum of these three squares is g(x) = 3x^2 + 2.
The allegation is equivalent to saying:
f(n) = 2(n9s)82(n0s)29
where the () denotes concatenation
f(n) corresponds to x = (n+1 9s)7, so f(0) is x = 97
g(97) = 3*97^2 + 2 = 28229
g(997) = 2982027
g(9997) = 299820027
etc
For an inductive proof, we already have the base case. I will work out going from k to k+1 using n=2 and n=3 since it is easier to represent with specific numbers. So this is admittedly a short cut and a less-than-vigorous proof.
Assume g(9997) = 299820027
Find g(99997) = g(90000 + 9997)
= 3(90000 + 9997)^2 + 2
= 3(8100000000 + 1799460000 + 9997^2) + 2
= 3(8100000000 + 1799460000) + 3*9997^2 + 2
= 24300000000 + 5398380000 + 299820027
= 29998200027
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Posted by Larry
on 2024-06-16 09:39:08 |
Inductive proof (not rigorous)