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

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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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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!)

--- SoupGate-Win32 v1.05
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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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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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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.

Your answer can be compressed by working mod 8 instead. Odd numbers all
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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"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.

(intelligent (or Clever) comments welcome!)

You're overcomplicating your reductio ad absurdem. Let S be the smallest
such power of 2. Your first steps show that S/4 must also satisfy the condition, therefore it wasn't the smallest.

Phil
--
We are no longer hunters and nomads. No longer awed and frightened, as we have gained some understanding of the world in which we live. As such, we can cast aside childish remnants from the dawn of our civilization.
-- NotSanguine on SoylentNews, after Eugen Weber in /The Western Tradition/

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