[Guest]

1) prove or disprove a^2 +/- b^2 +/- c^2 = any natural number (0 or more)

where a, b, c is any natural number

2) prove or disprove a^x +/- b^y = any natural number

where a, b is natural and x, y is natural > 1

[araver]

a^2 - b^2 - c^2 = any natural number.

since (n+1)^2-n^2-0^2=2*n+1 and (n+1)^2-n^2-1^2=2*n

Edited November 26, 2010 by araver

[Guest]

I don't get it...

What if a,b,c were 1,2,3 then a²-b²-c² = 1²-2²-3² = -12 ???

[araver]

I don't get it...

What if a,b,c were 1,2,3 then a²-b²-c² = 1²-2²-3² = -12 ???

that what the OP wants is to show that for every natural number there exists a pair (a,b,c) and a combination of + and - signs such that [...].

[araver]

Odd numbers and multiples of 4 are easy:

Odd numbers:

(n+1)^2-n^2=2*n+1

Even numbers that are also multiples of 4:

(n+2)^2-n^2=4*n+4=4*(n+1)

and 0^2-0^2=0 to take care of the last one

Even numbers that are not multiples of 4 (m = 4*p+2) are not that easy.

2 is possible (a=1,b=1 1+1=2)

I'm betting 6 isn't

a^x + b^y = 6 clearly has no solution (x,y>1)

a^x - b^y = 6 is much trickier. Haven't been able to find a proof.

[Guest]

Odd numbers and multiples of 4 are easy:

Odd numbers:

(n+1)^2-n^2=2*n+1

Even numbers that are also multiples of 4:

(n+2)^2-n^2=4*n+4=4*(n+1)

and 0^2-0^2=0 to take care of the last one

Even numbers that are not multiples of 4 (m = 4*p+2) are not that easy.

2 is possible (a=1,b=1 1+1=2)

I'm betting 6 isn't

a^x + b^y = 6 clearly has no solution (x,y>1)

a^x - b^y = 6 is much trickier. Haven't been able to find a proof.

(n+1)²-(n-1)² = 4n

Now for multiples of 2 that are not multiples of 4, I can tell you the two powers cannot be 2 and 2 cause:

(x+y)² - (x±z)² = (2y±2z)x + y²-z²

You'll need y+z to be add (otherwise 2y+2z will be divisible by 4) but for that you need one of them to be odd and the other even in which case y²-z² will be odd and you'll get an odd number

So we might need to go higher...

[araver]

I'm betting 6 isn't

a^x + b^y = 6 clearly has no solution (x,y>1)

a^x - b^y = 6 is much trickier. Haven't been able to find a proof.

I'm forced to stop as I feel it's way beyond my powers to prove a 40-year old open problem. I lack both the mathematical skills and the stubbornness required for that.

a^x - b^y = 6 is currently marked as an open question in number theory:

Sierpinski, W. – "250 problems in elementary number theory", 1970, problem no238a, p. 116.

Further open references here: Open Diophantine Equations, page 6.

Your question is both nice and devilish, phillip!

Edited November 27, 2010 by araver