Math Is Fun Forum

chris2108

Member

Registered: 2011-10-02

Posts: 1

Tough task

I'd be grateful for a solution

Find all pairs of positive itegers (x,y); x>0; y>0;

2^x + 5^y = c^2

c is integer

Bob

Administrator

Registered: 2010-06-20

Posts: 10,621

Re: Tough task

hi chris2108

Ouch, that is tough!  I take it you've found x = 2, y = 1.

After that it gets a lot harder.

Powers of 2 end in 2 or 4 or 8 or 6.

Powers of 5 end in 5.

So you can deduce that any possible c must end in 1 or 9.

Using that I tested all one or two digit possibilities for c and found no more solutions.

I'm leaning towards the idea that there is only the one solution but don't have an inkling of an idea how to prove it. 

Hopefully, wiser heads may jump in at this point with a number theory type answer.

Bob

later edit:

I have now tested power of 2 up to 4398046511104  and power of 5 up to 1220703125.

No new solutions!

Last edited by Bob (2011-10-02 12:11:05)

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

TMorgan

Member

Registered: 2011-04-13

Posts: 25

Re: Tough task

I got logic slightly different from Bob, though I don't have another answer either:

2^x ends in 2, 4, 6 or 8

5^y ends in 5

c^2 ends in 0, 1, 4, 5, 6 or 9

2^x + 5^y ends in 1, 3, 7 or 9

c^2 can't end in 3 or 7 so c^2 ends in 1 or 9

So c ends in 1, 3, 7 or 9.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,621

Re: Tough task

hi TMorgan

You said what I meant to say.  We are as one on this.

Bob

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Tough task

Hi;

Tested all numbers to

and

This could be extended but for one reason, there is what appears to be a good reason why there will not be any more solutions.

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In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.