[Guest]

Prove that there cannot exist a power of 2 that is the sum of three positive perfect squares.

[Guest]

I hope this is right :?

it is impossible because you can not split any power of 2 into thirds. the answer would be like 3.9999... or something like that

[Izzy]

I hope this is right :?

it is impossible because you can not split any power of 2 into thirds. the answer would be like 3.9999... or something like that

They wouldn't have to be even thirds. Like, 8, a power of 2, can be split into 1, 3, and 4.

It annoys me that I can't think of how to prove this. =/

Edited November 29, 2009 by Izzy

[Guest]

Dunno, working on it still, but mostly here is what I come up with.

(n-1)^2 + (n-3)^2 + (n-5)^2 < n^2 < (n-1)^2 + (n-3)^2 + (n-4)^2

example:

5^2 + 2^3 + 1^2 < 6^2 < 5^2 + 3^2 + 2^2

25 + 4 + 1 < 36 < 25 + 9 + 4

35 < 36 < 38

and that's as close as you will come to getting the numbers on either end for any value of n I think.

NM that doesn't hold true, but it'll be a statement somewhat like that.

Edited November 29, 2009 by Sanjurjo

[Guest]

oops i read the question wrong. lol my bad

Edited November 29, 2009 by StripesAndFins

[Guest]

Let 3 squares be a²+b²+c² = 2ⁿ

If a, b and c are all even then we can divide both sides by 4 continually until at least one of them is odd.

Let the new equation be:

x²+y²+z² = 2^m

If just one of x,y or z was odd (or all of them), then 2^m must be odd. 2^m = 1 clearly isn't a solution. The only possibility is that only one of the three are even. Let this one be x. I will now put x² = 4k, y = 2p+1, z = 2q+1

4k + (2p+1)² + (2q+1)² = 2^m

4k + (4p² + 4p + 1) + (4q² + 4p + 1) = 2^m

4(k + p² + p + q² + p) + 2 = 2^m

2(k + p² + p + q² + p) + 1 = 2^m-1

Therefore 2^m-1 is odd.

The only possibility is that 2^m-1 is 1 but that means that x²+y²+z² = 2 which has no solutions. Therefore there cannot be any squares x², y² & z² that sum a power of 2.

[Guest]

Since every positive number has a factor of 2, and not every positive number is a power of 2, when you add 3 squares of positive number together, it’s the same as multiply 3 by 2 initially, since the Square root of 3 is not an exact number, the product of 3 squares of positive even numbers cannot be a power of 2.

[Guest]

Since every positive number has a factor of 2, and not every positive number is a power of 2, when you add 3 squares of positive number together, it’s the same as multiply 3 by 2 initially, since the Square root of 3 is not an exact number, the product of 3 squares of positive even numbers cannot be a power of 2.

[Guest]

Suppose there did exist three positive integers, I,J and K so that

1) I² + J² + K² = 2^x for some integer x

First, we can rule out x being less than 4, because it is clear that no three positive perfect squares can add to 1,2,4 or 8. So x would have to be > 3.

A power of 2 which is 16 or greater must be 0 mod 16 (divisible by 16).

If an integer is a multiple of 4, its square is 0 mod 16,

(otherwise its square is 1,4 or 9 mod 16).

The only three of these that can add to 0 mod 16 is 0 + 0 + 0

So, I, J and K would all have to be multiples of 4, and their squares multiples of 16.

2) The only way three positive perfect squares can add to a power of two is if each square is a multiple of 16

But this runs into a problem. If we keep dividing each term of the equation by 16, one of the results must eventually be an odd number (the last stop is 1),

and hence not a multiple of 16 (a contradiction to #2 above)

So there cannot exist a power of 2 that is the sum of three positive perfect squares.

Edited November 29, 2009 by xamdam

[Guest]

Let 3 squares be a²+b²+c² = 2ⁿ

If a, b and c are all even then we can divide both sides by 4 continually until at least one of them is odd.

Let the new equation be:

x²+y²+z² = 2^m

If just one of x,y or z was odd (or all of them), then 2^m must be odd. 2^m = 1 clearly isn't a solution. The only possibility is that only one of the three are even. Let this one be x. I will now put x² = 4k, y = 2p+1, z = 2q+1

4k + (2p+1)² + (2q+1)² = 2^m

4k + (4p² + 4p + 1) + (4q² + 4p + 1) = 2^m

4(k + p² + p + q² + p) + 2 = 2^m

2(k + p² + p + q² + p) + 1 = 2^m-1

Therefore 2^m-1 is odd.

The only possibility is that 2^m-1 is 1 but that means that x²+y²+z² = 2 which has no solutions. Therefore there cannot be any squares x², y² & z² that sum a power of 2.

[unreality]

a^2 + b^2 + c^2 = 2^n

a,b,c, and n are all positive integers.

At the very lowest, if a=b=c=1, then their squares add to 3. Therefore n cannot be 0 or 1 and 2^n cannot be 1 or 2, and because 2^0 = 1 is the only odd positive power of two, 2^n must be an EVEN number.

that means a^2 plus b^2 plus c^2 is also even. That should mean that either all three are even or only one is, but let's double check:

a^2 + b^2 + c^2 = 2^n

EVEN + EVEN + EVEN = EVEN

EVEN + EVEN + ODD ≠ EVEN

EVEN + ODD + EVEN ≠ EVEN

ODD + EVEN + EVEN ≠ EVEN

EVEN + ODD + ODD = EVEN

ODD + EVEN + ODD = EVEN

ODD + ODD + EVEN = EVEN

ODD + ODD + ODD ≠ EVEN

Confirmed: all three squares must be even, or only one must be even.

If a square is even then its square root is also even, thus the even square is a multiple of 4. If all are even then you can just divide by 4 (and reduce the n in 2^n by two) until not so. The only times you can't do this are for n=0 and n=1, the lowest possible, and above we already proved that those are impossible.

Thus if it's possible to form the power of two from the three squares in the first place, if they are all even it means there is an even smaller version divided by four or perhaps sixteen or another power of four.

It's possible for EVEN EVEN EVEN squares to divide into EVEN ODD ODD squares which is the pattern we are looking for, ie...

4^2 + 6^2 + 10^2

--------------/4

=

2^2 + 3^2 + 5^2

in general, where k is any intenger, even perfect squares of the form (4k)^2 will divide by four to become (2k)^2 and thus also even; and even perfect squares of the form (4k+2)^2 will divide by four to become (2k+1)^2 and thus odd. So any one of the first type and two of the second type will form the EVEN ODD ODD square set.

In other words, we've reduced it down to one form: EVEN SQUARE + ODD SQUARE + ODD SQUARE = 2^n. If we can prove this to never exist, your statement is proved.

Three variables: x, y, z. They are all POSITIVE EVEN integers.

rewriting...

(x)^2 + (y-1)^2 + (z-1)^2 = 2^n

from there we manipulate algebraically...

x^2 + y^2 - 2y + 1 + z^2 - 2z +1 = 2^n

x^2 + y^2 + z^2 - 2y - 2z + 2 = 2^n

because x,y,z are mapped as positive even integers, if you divide them by two into g,h,i then g, h and i are all positive integers.

so we have

4g^2 + 4h^2 + 4i^2 - 4h - 4i + 2 = 2^n

now divide everything by two

2g^2 + 2h^2 + 2i^2 - 2h - 2i + 1 = 2^(n-1)

the number on the left has all even terms except for the 1 at the very end.

Thus 2^(n-1) must be odd, thus n-1 = 0 and n = 1

but we already proved this was impossible up above. There are no three positive squares that add to 2^1

It is impossible

[Guest]

a^2 + b^2 + c^2 = 2^l

Let's assume a, b, c are NOT all even, otherwise we could just divide by the power of 2 on both sides the equation

a, b, c couldn't be all odd, or the sum of the suqares would be odd

one has to be even while the other two are odd

suppose a is even

then a^2 is a multiple of 4

for b odd, b^2 divided by 4 willl give a remainder of 1

so does c^2

so a^2 + b^2 + c^2 will give a remainder of 2 when divided by 4, not 0 as if it's a power of 2.