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        riddles >> hard >> Sum Powers Of  2 And 5, Get Z
        
(Message started by: K Sengupta on Aug 6^th, 2007, 8:15am)

Title: Sum Powers Of 2 And 5, Get Z
Post by K Sengupta on Aug 6^th, 2007, 8:15am

Determine all possible non-negative integers (X, Y, Z) satisfying this equation:

2^X + 5^Y = 3^Z

Title: Re: Sum Powers Of 2 And 5, Get Z
Post by Eigenray on Aug 6^th, 2007, 10:38am

A start: note that z>0, and x>0 by parity. Suppose now that x>2 and y>0.

Mod 3, we see that [hide]x http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ne.gif y mod 2[/hide].

Mod 5, we see that [hide]x=z mod 2[/hide], either by enumerating the possibilities or because [hide](-1)^x=(2^x|5) = (3^z|5) = (-1)^z[/hide].

Mod 8, we see that [hide]y=z=0 mod 2[/hide].

Together we get a contradiction. So either:

(1) y=0, which gives 2^x + 1 = 3^z, which gives (x,z) = (1,1) or (3,2). [hide]If x>1, then z=2a is even, and 2^x = (3^a-1)(3^a+1) forces a=1.[/hide]

(2) x < 3, and y>0.

If x=2, then 4 + 5^y = 3^z, and mod 4, we see z=2k is even, and then 5^y = (3^k+2)(3^k-2). This is only possible if 3^k-2=1, 3^k+2=5, which gives the solution (x,y,z) = (2,1,2).

The remaining case is x=1, or

2 + 5^y = 3^z.

Mod 3, y=2k is even. This reduces to 5^k + ihttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif2 = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pm.gif(1+ihttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif2)^z when k is even, and 5^k + ihttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif2 = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pm.gif(1-ihttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif2)^z when k is odd.

Title: Re: Sum Powers Of 2 And 5, Get Z
Post by Eigenray on Aug 7^th, 2007, 7:12am

Ad hoc computer-assisted solution:

We have 2 + 5^y = 3^z.

[hideb]Suppose z>3. Working mod 3⁴, we find that y = 20 mod http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/phi.gif(3⁴). Thus y = 20 + 54k. Now 5⁵⁴=1 mod 109, so 3^z = 2+5²⁰ mod 109, which has no solutions. So z http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/le.gif 3.[/hideb]

It follows that the only solutions are (1,0,1), (3,0,2), (2,1,2), and (1,2,3).

Title: Re: Sum Powers Of 2 And 5, Get Z
Post by Hippo on Aug 7^th, 2007, 3:56pm

on 08/06/07 at 10:38:18, Eigenray wrote:

A start: note that z>0, and x>0 by parity. Suppose now that x>2 and y>0.

Mod 3, we see that [hide]x = -y mod 2[/hide].

Mod 5, we see that [hide]x=z mod 2[/hide], either by enumerating the possibilities or because [hide](-1)^x=(2^x|5) = (3^z|5) = (-1)^z[/hide].

Mod 8, we see that [hide]y=z=0 mod 2[/hide].

I get Mod 3: [hide]x+y=1 MOD 2[/hide].

Mod 5: [hide]x=z MOD 4[/hide].

Mod 8: [hide]y=z MOD 7[/hide].

Title: Re: Sum Powers Of 2 And 5, Get Z
Post by Eigenray on Aug 8^th, 2007, 12:04am

on 08/07/07 at 15:56:08, Hippo wrote:

I get Mod 3: [hide]x+y=1 MOD 2[/hide]

You're right, what I had typed was not what I meant.

Quote:

Mod 5: [hide]x=z MOD 4[/hide].

That should be [hide]x=-z MOD 4[/hide]; in particular [hide]x=z mod 2[/hide].

Quote:

Mod 8: [hide]y=z MOD 7[/hide].

I can't see how you're getting that. Mod 8, 5^y = 1,5,1,5,..., while 3^z = 1,3,1,3,.... So they are equal when y=z=0 mod 2.

Title: Re: Sum Powers Of 2 And 5, Get Z
Post by Hippo on Aug 9^th, 2007, 5:46am

Oops ... you are right ... I had find a mistake and made two others ... sorry once again ... I didn't spent a lot of time with the answer and it is not what I usually do.
:-[ Mod 5: It should be [hide]x+z=2 mod 4[/hide] ... ok ... .

Title: Re: Sum Powers Of 2 And 5, Get Z
Post by Eigenray on Aug 9^th, 2007, 5:59am

on 08/09/07 at 05:46:38, Hippo wrote:

Mod 5: It should be [hide]x+z=2 mod 4[/hide]

Mod 5, 3 = 2^-1, so 2^x = 3^z = 2^-z implies x=-z mod 4.

Title: Re: Sum Powers Of 2 And 5, Get Z
Post by Hippo on Aug 9^th, 2007, 6:28am

Ooops, oops :-[ I had solved 2^x+3^z=0 now. I should not write more about this topic ;)

Title: Re: Sum Powers Of 2 And 5, Get Z
Post by srn347 on Aug 26^th, 2007, 9:05pm

x, y, and z all equal infinity(positive or negative).

Title: Re: Sum Powers Of 2 And 5, Get Z
Post by Sameer on Aug 26^th, 2007, 9:12pm

on 08/26/07 at 21:05:54, srn347 wrote:

x, y, and z all equal infinity(positive or negative).

Infinity is not ~~a number~~ an integer

Edit: As pointed out by pex!! :)

Title: Re: Sum Powers Of 2 And 5, Get Z
Post by mikedagr8 on Aug 26^th, 2007, 9:45pm

Hahahaha, good call, you beat me to it. Infinity means, well, just wiktionary it, I could explain it in my own terms, except wiki does it better than me.

Title: Re: Sum Powers Of 2 And 5, Get Z
Post by pex on Aug 26^th, 2007, 11:37pm

on 08/26/07 at 21:12:36, Sameer wrote:

Infinity is not a number.

Of course it is - just not a real number, and most clearly not an integer, as this puzzle requires.

See Affinely Extended Real Numbers (http://mathworld.wolfram.com/AffinelyExtendedRealNumbers.html), Projectively Extended Real Numbers (http://mathworld.wolfram.com/ProjectivelyExtendedRealNumbers.html), among others.

Title: Re: Sum Powers Of 2 And 5, Get Z
Post by srn347 on Aug 27^th, 2007, 10:23am

Infinity is hyperreal. And how about x=1 y=2 z=3