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Mathematical Riddle

Here's a mathematical riddle that I was told on IRC (and was able to solve, after receiving a hint): find, without the aid of a computer, three natural numbers - x, y and z - so that x^2 + y^3 = z^5 (where "^" is power).

Feel free to post your solutions as comments below, just make sure you identify yourselves.

Comments

( 6 comments — Leave a comment )
adistav
Sep. 30th, 2007 06:15 am (UTC)
Solved the riddle, but I am curious what was the hint.
shlomif
Sep. 30th, 2007 08:59 am (UTC)
The Hint

The hint was: "choose x and y so that the '+' produces something nice.".

(Anonymous)
Sep. 30th, 2007 01:35 pm (UTC)
Well...
There is the trivial set of solutions, which is easy to put parametrically as 2^(m+2),2^(m+1),2^m for x,y,z respectively.

If we assume that x,y,z are 2^a, 2^b and 2^c, assuming both operands of the addition operation are equal (a.k.a. "the hint") and using the identity 2^n=2*2^(n-1) (which is the reason why this form of solution was chosen) we get the equations:

a+2=b+3=(c+5)-1

which have one degree of freedom as expressed above.

The assumptions limit the problem space, and I'm pretty sure other types of solutions can be derived, but it's not required by the problem and I'm tired now.

Was solved without use of the hint.

-- Arik
shlomif
Sep. 30th, 2007 06:09 pm (UTC)
Re: Well...


Hi Arik! You've got something a little wrong. The equation is:




2a == 3b == 5c-1




Because (2^a)^2 is 2^(a*2) not 2^(a+2). However, you can still use this.

(Anonymous)
Oct. 1st, 2007 04:07 am (UTC)
Trivial my hindside

This will teach not me to write trivial solutions at 5am. I didn't even check the result...

I'm embarrassed now.

Thanks,

-- Arik
shlomif
Dec. 17th, 2007 06:41 pm (UTC)
How I solved it.
Better late than never:

well, trying to get x^2 and y^3 to produce something nice, I was looking for a way to get x^2 and y^3 produce the same number so x^2+y^3 will be twice that. So I figured out I can have x=a^3 and y=a^2 and so eventually get 2*a^6 on the left hand side.

So now we have z^5 = 2 * a^6 = 2 * a * a^5. So z can be a * (2*a)^(1/5). So we can choose any a so that 2*a will be a power of 5. One example is a = 16, but there are also a = 4**5/2, a = 6**5/2, a = 8**5/2, etc.
( 6 comments — Leave a comment )

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