No triangular number, bigger than 1 is also a 4th power.
Is the above statement, attributed to (fill in):
a. true
b. false
c. not resolved yet
Please comment in details.
To resolve this, we need to form a Pellian equation:
Starting with n(n+1)/2 = x^4:
Say n is even, then for some a, a(2a+1) = x^4.
Since n and n+1 are coprime, we have a = p^4, (2a+1) = q^4, so that 2a=(q^4-1)and consequently q^4-2p^4 = 1.
Say n is odd, then for some b, (b-1)(2b) = x^4.
Since n and n+1 are coprime, we have (b-1) = p^4, (2b) = q^4, so that b = p^4+1, and consequently q^4-2p^4 = -1.
We now have the Pellian forms we needed, and Ljunggren, "Zur Theorie der Gleichung x^2+1=Dy^4" does the rest because he proves that apartr from 1, 169 is the only square (let alone higher powers) terms in the Pell sequence. In fact the closest we can get is q^2-2*169^2 = -1, with q = {239, -239}.
Edited on May 31, 2018, 10:38 pm
|
Posted by broll
on 2018-05-31 22:16:05 |