On Sunday, October 2, 2022 at 10:01:25 AM UTC-7, henh...@gmail.com wrote:
Can the sum of 2 (distinct, non-Zero) squares be a power of two ?
Can the sum of 3 (non-Zero) squares be a power of two ?
Can the sum of 4 (distinct, non-Zero) squares be a power of two ?
sum of 5 (distinct, non-Zero) squares
169 + 49 + 25 + 9 + 4 = 256
Can the sum of 3 (non-Zero) squares be a power of two ?i think i have a (reductio ad absurdum) Proof.
(thinking in Mod 4) Any square is congruent to ( ≡ ) 0 or 1.
4k + 1
4k + 2
4k + 3 --------- Can't be a power of 2.
So the three squares are (would have to be) all multiples of 4.
the sum S is expressible as 4 (a + b + c) = S
where a, b, c are also squares
(note : a, b, c are 1/2 of the original squares)
S is a power of 2. So (a + b + c) is also a power of 2.
The same argument and halving (1/2) operation. Repeat.
Eventually we'd get to (1 + B + C) = Power of 2
where B, C are squares
But this is impossible because
4k + 1
4k + 2
4k + 3 --------- Can't be a power of 2.
(intelligent (or Clever) comments welcome ! )
(note : a, b, c are 1/2 of the original squares) <--- Typo(note : a, b, c are 1/4 of the original squares)
Can the sum of 2 (distinct, non-Zero) squares be a power of two ?
Can the sum of 3 (non-Zero) squares be a power of two ?
Can the sum of 4 (distinct, non-Zero) squares be a power of two ?
sum of 5 (distinct, non-Zero) squares
169 + 49 + 25 + 9 + 4 = 256
Can the sum of 3 (non-Zero) squares be a power of two ?
On Sunday, October 2, 2022 at 10:23:50 AM UTC-7, henh...@gmail.com wrote:
On Sunday, October 2, 2022 at 10:01:25 AM UTC-7, henh...@gmail.com wrote:
Can the sum of 2 (distinct, non-Zero) squares be a power of two ?
Can the sum of 3 (non-Zero) squares be a power of two ?
Can the sum of 4 (distinct, non-Zero) squares be a power of two ?
sum of 5 (distinct, non-Zero) squares
169 + 49 + 25 + 9 + 4 = 256
Can the sum of 3 (non-Zero) squares be a power of two ?i think i have a (reductio ad absurdum) Proof.
(thinking in Mod 4) Any square is congruent to ( ≡ ) 0 or 1.
4k + 1
4k + 2
4k + 3 --------- Can't be a power of 2.
So the three squares are (would have to be) all multiples of 4.
the sum S is expressible as 4 (a + b + c) = S
where a, b, c are also squares
(note : a, b, c are 1/2 of the original squares)
S is a power of 2. So (a + b + c) is also a power of 2.
The same argument and halving (1/2) operation. Repeat.
Eventually we'd get to (1 + B + C) = Power of 2
where B, C are squares
But this is impossible because
4k + 1
4k + 2
4k + 3 --------- Can't be a power of 2.
(intelligent (or Clever) comments welcome ! )(note : a, b, c are 1/4 of the original squares)
(note : a, b, c are 1/2 of the original squares) <--- Typo
this proof is better expressed in terms of a [ reduced-Form ] equation
where it's defined as ... (e.g.)
169 + 49 + 25 + 9 + 4 = 256 ( reduced-Form )
multiply all terms by 4 -------- ( not reduced-Form )
and .............
On Sunday, October 2, 2022 at 10:54:41 AM UTC-7, henh...@gmail.com wrote:
On Sunday, October 2, 2022 at 10:23:50 AM UTC-7, henh...@gmail.com wrote: >>> On Sunday, October 2, 2022 at 10:01:25 AM UTC-7, henh...@gmail.com wrote: >>>> Can the sum of 2 (distinct, non-Zero) squares be a power of two ?
(note : a, b, c are 1/4 of the original squares)
Can the sum of 3 (non-Zero) squares be a power of two ?
Can the sum of 4 (distinct, non-Zero) squares be a power of two ?
sum of 5 (distinct, non-Zero) squares
169 + 49 + 25 + 9 + 4 = 256
Can the sum of 3 (non-Zero) squares be a power of two ?i think i have a (reductio ad absurdum) Proof.
(thinking in Mod 4) Any square is congruent to ( ≡ ) 0 or 1.
4k + 1
4k + 2
4k + 3 --------- Can't be a power of 2.
So the three squares are (would have to be) all multiples of 4.
the sum S is expressible as 4 (a + b + c) = S
where a, b, c are also squares
(note : a, b, c are 1/2 of the original squares)
S is a power of 2. So (a + b + c) is also a power of 2.
The same argument and halving (1/2) operation. Repeat.
Eventually we'd get to (1 + B + C) = Power of 2
where B, C are squares
But this is impossible because
4k + 1
4k + 2
4k + 3 --------- Can't be a power of 2.
(intelligent (or Clever) comments welcome ! )
(note : a, b, c are 1/2 of the original squares) <--- Typo
this proof is better expressed in terms of a [ reduced-Form ] equation
where it's defined as ... (e.g.)
169 + 49 + 25 + 9 + 4 = 256 ( reduced-Form )
multiply all terms by 4 -------- ( not reduced-Form )
and .............
this same argument (proof) wroks for...
Can the sum of 2 (distinct, non-Zero) squares be a power of two ?
and a similar argument (proof) wroks for...
Can the sum of 4 (distinct, non-Zero) squares be a power of two ?
4 squares that are all odd (i.e. = 1 mod 4) also appears to be
impossible, though the proof is a little more involved.
On Sunday, October 2, 2022 at 10:01:25 AM UTC-7, henh...@gmail.com wrote:
Can the sum of 3 (non-Zero) squares be a power of two ?
i think i have a (reductio ad absurdum) Proof.
(thinking in Mod 4) Any square is congruent to ( ≡ ) 0 or 1.
4k + 1
4k + 2
4k + 3 --------- Can't be a power of 2.
So the three squares are (would have to be) all multiples of 4.
the sum S is expressible as 4 (a + b + c) = S
where a, b, c are also squares
(note : a, b, c are 1/2 of the original squares)
S is a power of 2. So (a + b + c) is also a power of 2.
The same argument and halving (1/2) operation. Repeat.
Eventually we'd get to (1 + B + C) = Power of 2
where B, C are squares
But this is impossible because
4k + 1
4k + 2
4k + 3 --------- Can't be a power of 2.
(intelligent (or Clever) comments welcome!)
Sysop: | Keyop |
---|---|
Location: | Huddersfield, West Yorkshire, UK |
Users: | 322 |
Nodes: | 16 (2 / 14) |
Uptime: | 130:36:57 |
Calls: | 7,067 |
Calls today: | 4 |
Files: | 12,493 |
Messages: | 5,499,499 |