[Guest]

ok so, theres a number sequence that goes a little something like this

xⁿ + yⁿ = zⁿ

Now, the x, y, and Z can be any number you want. The trick is that the n facotr has to be 2 or greater. I know of two sets of numbers that will work. Im only going to give you one set though. who can find the other one that I know. Also, who can think of anymore?

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x, y, and z are as follows

3² + 4² = 5²

9 + 16 = 25

25 = 25

[Guest]

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20²+21²=29²

[fabpig]

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12^2 + 5^2 = 13^2

144 + 25 = 169

and multiples

Edited November 9, 2010 by fabpig

[Guest]

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12²+5²=13²

There are many others with n=2 since that is the pythagorean theorem. I'm sure there are others that work with n>2.

Edited November 9, 2010 by lsmithers

[Guest]

If x,y and z can be "any number you want", then they don't have to be integers.

So I can pick any n,x and y, then calculate z (which probably won't be an integer).

From there I can get an infinite number of solutions.

[witzar]

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for n=2:

http://en.wikipedia.org/wiki/Pythagorean_triple

for n>2: http://en.wikipedia.org/wiki/Fermat%27s_Last_Theorem

[Guest]

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If you have one solution, an infinite number of solutions can be derived by just multiplying x, y, and z by the same number. Even without this trick, I believe there are an infinite number of solutions that are not multiples of each other, and as far back as Euclid there was a formula for generating these numbers.

For n>2, Fermat's last theorem shows that there are no solutions.

Edited November 9, 2010 by rajat_magic

[Guest]

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33² + 44² = 55²

1089 + 1936 = 3025

3025 = 3025

333² + 444² = 555²

110889 + 197136 = 308025

308025 = 308025

Edited November 9, 2010 by B0SS_LADiEE

[Guest]

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just keep the exponent at two and multiply the other numbers by anything.

Infinite amount of answers

  On 11/9/2010 at 5:59 PM, rajat_magic said:

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If you have one solution, an infinite number of solutions can be derived by just multiplying x, y, and z by the same number. Even without this trick, I believe there are an infinite number of solutions that are not multiples of each other, and as far back as Euclid there was a formula for generating these numbers.

For n>2, Fermat's last theorem shows that there are no solutions.

[Guest]

let n =1, then you really have infinite solutions!

[EventHorizon]

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Since you say the numbers can be any number you want, why not negatives? Duplicates?

Therefore, any solution of the following forms would work:

1. x=-y,z=0,n=odd number

2. x=z,y=0,n=any number (excluding 0 due to 0^0, but you said n>=2 anyway)

"Any number you want" could also include real or complex numbers, which kills the puzzle (e.g., you can just calculate the remaining value).

But, I'm pretty sure you meant to restrict to positive integers, in which case you get Pythagorean triples and Fermat's last theorem (which have been listed in responses already).

[Guest]

It is the easier riddle ever.We all (?) know that in an right triangle this happens:

...|\

...|..\ ......................a² = b² +c²

...|....\

...|......\

b.|....... \

...|..........\......a

...|............\

...|..............\

...|................\

...|__________ \

............ c

We can set any 2 numbers we want and do the magic: (Parameter: a > b and a > c)

e.g : Set c = 8 , b = 3 => a² = 3² + 8² => a² = 9 + 64 => a² = 73 => a = √73 => a = 8,54...(almost)

so

(almost)8,54² = 3² + 8² => 73 = 9 + 64 => 73 = 73

e.g : Set a = 40 , c = 2 => 40² = b² + 2² => b² = 40² - 2² => b² = 1600- 4 => b² = 1596 => b = √1596 => b = 39, 94(almost)

39,94² =... (Same with above, no real reason to show)

So here you are.Answers? If we set the possible solutions as K and all the right triangles combinations as J then:

K = J (Probably Unlimited)

*The square root of a number is +- √x but there is not such thing as negative distance

*We can also add the possiblity of negative numbers.However this is not possible on a triangle becouse there is no negative

distance but becouse the numbers will be x^2 they will always give possitive result.The parameter then would be

(Parameter: |a|>|b| and |a|>|c|)

However there is no need to do that.We can all understand that using the triangle property we can have unlimited combinations, no reason to say all ways to solve this puzzle.

Edited November 10, 2010 by Tsopi

[Guest]

  On 11/10/2010 at 8:40 PM, Tsopi said:

It is the easier riddle ever.

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We all (?) know that in an right triangle this happens:

...|\

...|..\ ......................a² = b² +c²

...|....\

...|......\

b.|....... \

...|..........\......a

...|............\

...|..............\

...|................\

...|__________ \

............ c

We can set any 2 numbers we want and do the magic: (Parameter: a > b and a > c)

e.g : Set c = 8 , b = 3 => a² = 3² + 8² => a² = 9 + 64 => a² = 73 => a = √73 => a = 8,54...(almost)

so

(almost)8,54² = 3² + 8² => 73 = 9 + 64 => 73 = 73

e.g : Set a = 40 , c = 2 => 40² = b² + 2² => b² = 40² - 2² => b² = 1600- 4 => b² = 1596 => b = √1596 => b = 39, 94(almost)

39,94² =... (Same with above, no real reason to show)

So here you are.Answers? If we set the possible solutions as K and all the right triangles combinations as J then:

K = J (Probably Unlimited)

*The square root of a number is +- √x but there is not such thing as negative distance

*We can also add the possiblity of negative numbers.However this is not possible on a triangle becouse there is no negative

distance but becouse the numbers will be x^2 they will always give possitive result.The parameter then would be

(Parameter: |a|>|b| and |a|>|c|)

However there is no need to do that.We can all understand that using the triangle property we can have unlimited combinations, no reason to say all ways to solve this puzzle.

[dark_magician_92]

  On 11/9/2010 at 5:05 PM, jakeknowslittle said:

ok so, theres a number sequence that goes a little something like this

xⁿ + yⁿ = zⁿ

Now, the x, y, and Z can be any number you want. The trick is that the n facotr has to be 2 or greater. I know of two sets of numbers that will work. Im only going to give you one set though. who can find the other one that I know. Also, who can think of anymore?

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x, y, and z are as follows

3² + 4² = 5²

9 + 16 = 25

25 = 25

Edited January 7, 2011 by dark_magician_92

[Guest]

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take any two integers: a and b such that b>a

x,y, and z are then: 2*a*b, b^2-a^2, and b^2+a^2 respectively