Three Magical Numbers
There are three simple numbers which has a mysterious property. Sum of any two numbers is a perfect square. Can you tell the numbers ?
Labels: mathschallenge
posted by ToM at
7:14 AM
Wednesday, August 19, 2009
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36 Comments:
2,2,2...not a guess but calculation using the fact that the only simple number divideable by 2 is 2:D
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Very nice answer. I hadn't noticed that one. I don't understand your explanation though.
Now I see 8,8,8; 18,18,18 etc.
Or -1,0,1 as a solution in (Gaussian) integers. if allow 0 as being a square.
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milos, take a look at the weight problem, I've updated it, skip to the last few posts.
Also 1,8,8.
Also 7,2,2; 7,18,18; 4,32,32; 14,50,50 etc. This is fun.
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I think tom means 3 different numbers, nobody said positive numbers =)
[7,-3,2]
for me these are all perfect squares.
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t:b:.:H - Very clever - I do like that.
Despite the fun, I think that the only interesting bit left to do, is to show it can't be done with real distinct positive integers - I fancy that is implied in the challenge.
I'm not saying the other stuff is off topic by any means, though.
sorry english is not my mother language but does simple number mean a number divideable only by himslef and 1???
@Milos, thats a Prime Number
assuming that all numbers have to be different however don't have to be positive, isn't there an infinite amount of solutions to this?
cause this is wat I got:
pick 3 different numbers, label them x,y,z and follow this process:
((x^2)+(y^2)+(z^2))/2 = A
therefore: B=(x^2)-A
and C=(y^2)-B
where A,B,C is your solutions...
substituting 1,2,3
I got 3,-2,6
I just realised, there are conditions to which numbers you chose for x,y,z....so far i see that if u keep x and 7 odd, and y even...you'll always come up with a rational number...but I'm in a rush now...cya
EDIT:
"if you keep x and z odd, and y even."
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sorry to burst a bubble, but 7;-3;2 wont work because the square root of -1 is imaginary...
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30;-5; and 6 :D
oh, and sorry, i wasnt paying enough attention...
oh, finally, one without a negative [0;16;9]
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Anonymous 7:41, that's OK, "the man who made no mistakes made nothing". Your answer is good with me.
Anonymous 7:49. Oh yes, and there's plenty more like that too.
Looks like everyone's finding cheeky but valid solutions. I notice no one has the the hard one with 3 distinct +ve integer and pucker perfect squares, or even 3 distinct +ve reals and pucker perfect squares.
0 and the squares of the two smaller numbers of any Pythagorean triple.
its very simple.
if require positive no.s,the no.s are 1,24,120
Last Anonymous, that's what I reckoned.
6,19,30; I'm personally relieved I hadn't published a proof that
that solution was impossible.
Lara had made a typo, she should have written:
(-x^2 + y^2 + z^2)/2 = A.
I've slightly tweaked her definitions for aesthetic reasons.
A selection rule for x,y,z follows from the need for (...)/2 to be
integral. x,y,z all 3 even or 1 even and 2 odd.
The rest is correct, i.e. B = z^2 - A, C = x^2 - B.
x,y,z = 1,2,3 => x^2,y^2,z^2 = 1,4,9 and A,B,C = 6,3,-2
x,y,z = 5,6,7 => x^2,y^2,z^3 = 25,36,49 and A,B,C = 30,19,6
If you prefer: B = (x^2 - y^2 + z^2)/2 and C = (x^2 + y^2 - z^2)/2
You can now rattle of solutions. Thanks Lara.
I'm feeling foolish for not cracking that one immediately.
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Just for neatness: define S = (x^2 + y^2 + z^2)/2, then
A = S - x^2, B = S - y^2, C = S - z^2
If drop assumption that A,B,C are integral,
then e.g. A,B,C = 9.5, 6.5, -5.5 does it.
A,B,C can be guaranteed +ve if choose z>y>x and x^2 + y^2 > z^2
Although the problem was stated in terms of squares, it also works
for cubes and up - with the obvious changes to the equations.
i believe the answer is 1,3,5
That doesn't work. 1+3 = 4 is OK
1+5=6 and 3+5=8 aren't perfect squares.
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I'm such a pedant.
The a,b,c all +ve condition should have started: z>y>x>0
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If there's anybody out there?(good song)
The conditions on x,y,z corresponds to them being able to form the sides of a triangle.
...I thought not
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