Numbers p; p = (p+1)^k + (p+1)^k – (p+2)^k

I’ll be looking for numbers p such that p = (p + 1)^k + (p + 1)^k – (p + 2)^k

If k = 2, then
p = (p + 1)^2 + (p + 1)^2 – (p + 2)^2
-p^2 + p + 2 = 0
Solutions: p = -1     p = 2

2 = 3 + 3 – 4    and    2 = 3^2 + 3^2 – 4^2

If k = 3, then
p = (p + 1)^3 + (p + 1)^3 – (p + 2)^3
-p^3 + 7p + 6 = 0
Solutions: p = -2     p = -1     p = 3
The only acceptable solution: p = 3

3 = 4 + 4 – 5    and    3 = 4^3 + 4^3 – 5^3

For any k > 3, there does not exist an integer p such that
p = (p + 1)^k + (p + 1)^k – (p + 2)^k

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About benvitalis

math grad - Interest: Number theory
This entry was posted in Math Beauty and tagged , . Bookmark the permalink.

1 Response to Numbers p; p = (p+1)^k + (p+1)^k – (p+2)^k

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