• #### Can the sum of 2 (or 3 or 4) squares be a power of two ?

From henhanna@gmail.com@21:1/5 to All on Sun Oct 2 10:01:23 2022
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

--- SoupGate-Win32 v1.05
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• From henhanna@gmail.com@21:1/5 to henh...@gmail.com on Sun Oct 2 10:54:40 2022
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 1/2 of the original squares) <--- Typo
(note : a, b, c are 1/4 of the original squares)

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 .............

--- SoupGate-Win32 v1.05
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• From henhanna@gmail.com@21:1/5 to henh...@gmail.com on Sun Oct 2 10:23:48 2022
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.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From henhanna@gmail.com@21:1/5 to henh...@gmail.com on Sun Oct 2 12:52:51 2022
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 ?

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)

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 ?

so this is nice... i feel like a boy who has learned how to ride a bicycle.

a better analogy would be....

how i felt when i learned how to use all the Buttons on my simple calculator.
%, AC, C, and the Memory buttons (MC, MR, M-, M+)

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• From Edward Murphy@21:1/5 to henh...@gmail.com on Sun Oct 2 14:38:16 2022
On 10/2/2022 12:52 PM, henh...@gmail.com wrote:

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 ?

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)

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 ?

It appears to be sound for 2 or 3 squares, as well as 4 squares if
they're all even (i.e. = 0 mod 4).

4 squares that are all odd (i.e. = 1 mod 4) also appears to be
impossible, though the proof is a little more involved.

a = 2e + 1 a^2 = 4e^2 + 4e + 1
b = 2f + 1 b^2 = 4f^2 + 4f + 1
c = 2g + 1 c^2 = 4g^2 + 4g + 1
d = 2h + 1 d^2 = 4h^2 + 4h + 1

a^2 + b^2 + c^2 + d^2 = 4(e^2 + f^2 + g^2 + h^2 + e + f + g + h + 1)

If the right side is a power of 2, then
x = e^2 + f^2 + g^2 + h^2 + e + f + g + h + 1
must also be a power of 2. But it can't, because:

x = e(e+1) + f(f+1) + g(g+1) + h(h+1) + 1

e(e+1) is even (because either e is even, or e is odd and e+1 is even)
f(f+1) is even
g(g+1) is even
h(h+1) is even

Without loss of generality, let
a > b > c > d > 0
which also gives
e > f > g > h >= 0

So x is the sum of four even numbers (at most one of which may equal
zero, the other three are positive) and 1, which can't be a power of 2.

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• From Gareth Taylor@21:1/5 to emurphy42@zoho.com on Mon Oct 3 23:24:08 2022
In article <thd0c9\$pq\$1@gioia.aioe.org>,
Edward Murphy <emurphy42@zoho.com> wrote:

4 squares that are all odd (i.e. = 1 mod 4) also appears to be
impossible, though the proof is a little more involved.

square to 1 mod 8, and so four of them can't sum to a power of 2 unless
we're looking at 2^2.

This also extends to information about the five squares case. If all
five are even then we can cancel down. So assume at least one is odd.
Noting that even numbers square to 0 or 4 mod 8, we must then have four
odds and one number whose square is 4 mod 8. I.e., four odds and one
number which is 2 mod 4, like in the example of the the first post.

Gareth

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• From Phil Carmody@21:1/5 to henh...@gmail.com on Thu Oct 6 15:55:10 2022
"henh...@gmail.com" <henhanna@gmail.com> writes:
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.