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fatball
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Perfect squares
« on: Jan 22nd, 2006, 8:51pm » |
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Find all pairs of positive integers, x, y, such that x2 + 3y and y2 + 3x are both perfect squares.
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Eigenray
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Re: Perfect squares
« Reply #1 on: Jan 23rd, 2006, 3:23am » |
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Nice problem. I guess I won't be getting any sleep before the semester starts this morning.
| hidden: | Say x < y. Now, for some r>0, we have
y2 + 3x = (y+r)2 = y2 + 2ry + r2,
or 3x=2ry+r2, and since x<y we must have r=1, 3x=2y+1. So we may write
m2 = x2 + 3y = x2 + 3(3x-1)/2,
16m2 = 16x2 + 72x - 24 = (4x+9)2 - 105,
3*5*7 = 105 = (4x+9)2-16m2 = (4x+9+4m)(4x+9-4m),
so (4x+9-4m) = d, (4x+9+4m) = 105/d, for d=1,3,5, or 7 (taking m>0). We can check these directly, or note that
d + 105/d = 2(4x+9) = 2 mod 8 implies d=1 mod 4,
but in any case the only possibilities are d=1, x=11, or d=5, x=1. Recalling 3x=2y+1, this gives (x,y) = (11, 16) or (1, 1) as the only pairs with x<y. |
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Barukh
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Re: Perfect squares
« Reply #2 on: Jan 23rd, 2006, 4:38am » |
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on Jan 23rd, 2006, 3:23am, Eigenray wrote:| I guess I won't be getting any sleep before the semester starts this morning. |
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Does that mean we will be missing you frequently in the coming months? 
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Eigenray
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Re: Perfect squares
« Reply #3 on: Jan 24th, 2006, 6:08am » |
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It means next time I stay up all night doing math, it'll probably be for a course.
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fatball
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Re: Perfect squares
« Reply #4 on: Jan 24th, 2006, 10:31am » |
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Eigenray, thanks for your effort.  Your answer is near perfect except that there is no need to assume x<y as they are symmetric equations. That is also why you missed another pair of solution: (16,11)
Here is another solution:
As x and y are positive integers, then we can write x2 + 3y and y2 +3x in the form (x + a)2 and (y + b)2 respectively where a, b are also positive integers.
Expanding and eliminating the squared terms, we get 2 linear simultaneous equations:
3y = 2ax + a2
3x = 2by + b2
Solving, we get
x = (2a2b + 3b2)/(9-4ab)
y = (2b2a + 3a2)/(9-4ab)
Since a and b are positive, the numerators in the above fractions are positive; for the denominators to be positive (in order to make x and y positive), we must therefore have ab = 1 or 2.
Possibilities are (a,b) = (1,1), (1,2), (2,1) => (x,y) = (1,1), (16,11), (11,16).
Generalization: If there is no restriction on the sign of x and y (still integers), is there any additional solution?
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| « Last Edit: Jan 24th, 2006, 10:33am by fatball » |
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JocK
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Re: Perfect squares
« Reply #5 on: Jan 24th, 2006, 10:58am » |
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on Jan 24th, 2006, 10:31am, fatball wrote:Eigenray, thanks for your effort. Your answer is near perfect except that there is no need to assume x<y as they are symmetric equations. That is also why you missed another pair of solution: (16,11)
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Isn't it obvious that Eigenray had this in mind when he wrote "Say x < y" and "this gives (x,y) = (11, 16) or (1, 1) as the only pairs with x<y"..?
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solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it.
xy - y = x5 - y4 - y3 = 20; x>0, y>0.
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fatball
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Re: Perfect squares
« Reply #6 on: Jan 24th, 2006, 12:02pm » |
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Well yes, his answer is perfect based on his assumption but it does not mean that it is a complete answer given the unneccessary assumptions...
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| « Last Edit: Jan 24th, 2006, 12:04pm by fatball » |
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JocK
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Re: Perfect squares
« Reply #7 on: Jan 24th, 2006, 12:08pm » |
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Eigenray did not assume x < y. Rather, being aware of the permutation symmetry, he realised he only needed to investigate one of the two cases x<y and y<x.
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| « Last Edit: Jan 24th, 2006, 12:10pm by JocK » |
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solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it.
xy - y = x5 - y4 - y3 = 20; x>0, y>0.
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fatball
Senior Riddler
Can anyone help me think outside the box please?
Gender:
Posts: 315
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Re: Perfect squares
« Reply #8 on: Jan 24th, 2006, 12:16pm » |
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Point taken.
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