Shaun Isherwood
@shaunlovesmaths.bsky.social
I'm a PhD student at the University of Aberdeen studying group theory, an area of mathematics concerning symmetry. I have schizophrenia.
I'm wrapping up my research project.
I'm wrapping up my research project.
(I've already emailed my supervisor about this today, so there's no need to suggest I do that.)
January 11, 2026 at 11:11 PM
(I've already emailed my supervisor about this today, so there's no need to suggest I do that.)
December 30, 2025 at 12:31 AM
Is this modulo one?
December 23, 2025 at 1:51 PM
Is this modulo one?
Here's a proof using complex numbers:
n⁴+4
=(n²+2i)(n²-2i)
=(n+1+i)(n-1-i)(n+1-i)(n-1+i)
=(n+1+i)(n+1-i)(n-1-i)(n-1+i)
=((n+1)²+1)((n-1)²+1),
where the last equality is by a difference of squares.
Note that each factor is greater than 1 when n>1. Q.E.D.
n⁴+4
=(n²+2i)(n²-2i)
=(n+1+i)(n-1-i)(n+1-i)(n-1+i)
=(n+1+i)(n+1-i)(n-1-i)(n-1+i)
=((n+1)²+1)((n-1)²+1),
where the last equality is by a difference of squares.
Note that each factor is greater than 1 when n>1. Q.E.D.
December 14, 2025 at 12:29 PM
Here's a proof using complex numbers:
n⁴+4
=(n²+2i)(n²-2i)
=(n+1+i)(n-1-i)(n+1-i)(n-1+i)
=(n+1+i)(n+1-i)(n-1-i)(n-1+i)
=((n+1)²+1)((n-1)²+1),
where the last equality is by a difference of squares.
Note that each factor is greater than 1 when n>1. Q.E.D.
n⁴+4
=(n²+2i)(n²-2i)
=(n+1+i)(n-1-i)(n+1-i)(n-1+i)
=(n+1+i)(n+1-i)(n-1-i)(n-1+i)
=((n+1)²+1)((n-1)²+1),
where the last equality is by a difference of squares.
Note that each factor is greater than 1 when n>1. Q.E.D.
Assuming the positive square root, the answer is 27.
Use the difference of two squares to get the square root of (243×3), which is 27 (by breaking 243 into into prime decomposition; it is a power of 3).
Easy.
Use the difference of two squares to get the square root of (243×3), which is 27 (by breaking 243 into into prime decomposition; it is a power of 3).
Easy.
September 27, 2025 at 10:59 AM
Assuming the positive square root, the answer is 27.
Use the difference of two squares to get the square root of (243×3), which is 27 (by breaking 243 into into prime decomposition; it is a power of 3).
Easy.
Use the difference of two squares to get the square root of (243×3), which is 27 (by breaking 243 into into prime decomposition; it is a power of 3).
Easy.