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Topic: ALWAYS A PERFECT SQUARE (Read 1553 times) |
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pcbouhid
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ALWAYS A PERFECT SQUARE
« on: Nov 29th, 2005, 8:23am » |
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Find four natural numbers such that the square of each of them, when added to the sum of the remaining three, agains yields a perfect square.
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Barukh
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Re: ALWAYS A PERFECT SQUARE
« Reply #1 on: Nov 29th, 2005, 8:52am » |
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Paolo, the trivial answer would be 1, 1, 1, 1, but that's probably not what you want. 
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JocK
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Re: ALWAYS A PERFECT SQUARE
« Reply #2 on: Nov 29th, 2005, 10:22am » |
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Extending Barukh's solution, an infinite number of (slightly) less trivial answers is given by:
any (1, k, k, k) with 3k+1 a perfect square.
k = n(3n-2) : k = 1, 8, 21, 40, ...
or:
k = n(3n+2) : k = 5, 16, 33, 56, ...
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| « Last Edit: Nov 29th, 2005, 11:59am 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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JohanC
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Re: ALWAYS A PERFECT SQUARE
« Reply #3 on: Nov 29th, 2005, 1:02pm » |
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Two other solutions, not including the number 1:
6 6 11 11
40 57 96 96
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JocK
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Re: ALWAYS A PERFECT SQUARE
« Reply #4 on: Nov 29th, 2005, 1:44pm » |
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Would there be a solution with four distinct natural numbers?
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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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SMQ
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Re: ALWAYS A PERFECT SQUARE
« Reply #5 on: Nov 29th, 2005, 2:05pm » |
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Not with the largest number < 2000, which is as far as my exhaustive computerized search has reached so far...
[edit]Nor with the largest number less than 3000. Unless someone has a better strategy than searching all a < b < c < d, that's as far as I have time to look, as I need my computer back for work I actually get paid for...[/edit]
--SMQ
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| « Last Edit: Nov 30th, 2005, 5:39am by SMQ » |
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--SMQ
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pcbouhid
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Re: ALWAYS A PERFECT SQUARE
« Reply #6 on: Nov 30th, 2005, 10:16am » |
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In short, with no explanation (that I´ll be glad to post if you want):
Let the numbers be x, y, z, and u.
1) The numbers are all distinct: this case is not possible.
2) Precisely two of the integers are equal: x = y = 96, z = 57, u = 40.
3) The numbers are two pairs of equal numbers: x = y = 11, z = u = 6.
4) Three of the numbers are equal: u = 1, and x = y = z = k(3k+2) or k(3k-2), k an arbitrary integer.
5) All the numbers are the same: x = y = z = u = 1.
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JocK
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Re: ALWAYS A PERFECT SQUARE
« Reply #7 on: Nov 30th, 2005, 10:43am » |
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Do you mean to say that we found all integer solutions..?
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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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pcbouhid
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Re: ALWAYS A PERFECT SQUARE
« Reply #8 on: Dec 1st, 2005, 6:02am » |
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Exactly!!!!
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JocK
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Re: ALWAYS A PERFECT SQUARE
« Reply #9 on: Dec 1st, 2005, 9:25am » |
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Can you prove that?
(Feels good... being given the opportunity for bouncing back a question ..  )
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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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pcbouhid
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Re: ALWAYS A PERFECT SQUARE
« Reply #10 on: Dec 2nd, 2005, 5:51am » |
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As I stated, I´ll be glad to.
Just give me a little time to type it. I have to solve some things out there. Back again (about 2 hours) I´ll post the whole explanation. And a very nice one in the "hard" section.
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pcbouhid
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Re: ALWAYS A PERFECT SQUARE
« Reply #11 on: Dec 2nd, 2005, 9:28am » |
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Detailed solution:
The problem gives rise to the following system of equations to be solved in integers (where x, y, z, and u are the integers sought:
x^2 + y + z + u = (x + v)^2
y^2 + x + z + u = (y + w)^2
z^2 + x + y + u = (z + t)^2
u^2 + x + y + z = (u + s)^2
or
y + z + u = 2vx + v^2
x + z + u = 2wy + w^2
x + y + u = 2tz + t^2
z + y + z = 2su + s^2............(eqs. 1)
If we add these equations, we obtain:
(2v - 3)x + (2w - 3)y + (2t - 3)z + (2s - 3)u + v^2 + w^2 + t^2 + s^2 = 0...........(eq. 2)
Note that in eq. 2 at least one of the numbers (2v - 3), (2w - 3), (2t - 3), (2s - 3) is negative; otherwise there would be on the left of the eq. the sum of positive integers.
Let us assume that (2v - 3) < 0. This is possible only if v = 0 or v = 1. In the first event, the first eqs. of system (1) yields y + z + u = 0, which is untenable for y, z, u all positive. Hence it must be assumed that all the integers v, w, t, and s are positive, and that v = 1. In this event eq. 2 can be rewritten in the form
x = (2w - 3)y + (2t - 3)z + (2s - 3)u + w^2 + t^2 + s^2 + 1.........(eq. 3)
We now consider the several posibilities.
I) The numbers x, y, z, u are all distinct.
Here, the integers v, w, t, s are also all different; if, for example, v = w, the difference of the first two of the eqs. in (1) yields (y - x) = 2v(x - y), which is impossible for positive v and x != y. Further, if we assume v = 1, the first of equations (1) yields 2x = y + z + u - 1, x = 1/2y + 1/2z + 1/2u - 1/2, which is inconsistent with eq. 3, where the coef. of y, z, u in the right member are positive integers (since w, t, or s cannot be equal to 1 because they are distinct from v wich is equal to 1). Therefore, this case is not possible.
II) Precisely two of the integers x, y, z, u are equal.
Here we must separately investigate two cases.
a) If z = u, then t = s. Eq. 3 and the eqs. 1 now yield:
x = (2w - 3)y + 2(2t - 3)z + w^2 + 2t^2 + 1
2x = y + 2z - 1.
As before, these eqs. are inconsistent.
b) If x = y, then w = v = 1. Eq. 2 and the eqs. 1 yield, respectively:
2x = (2t - 3)z + (2s - 3)u + t^2 + s^2 + 2
x = z + u - 1
Substituting the second equality in the first:
(2t - 5)z + (2s - 5)u + t^2 + s^2 + 4 = 0.....eq. 4
from which it follows that at least one of the members (2t - 5) or (2s - 5) must be negative.
Assume (2t - 5) < 0; since t > 0 and t != 1 (for v = 1, t != v, since z != x), it follows that t = 2. Now, if twice the first of the eqs. (1) is added to the third equation, we obtain
4z + 4x + 6 = 4x + 2z + 3u,
that is, z = 3u/2 - 3. Substituting this into eq. 4, along with t = 2, we have:
(4s - 13)u + 2s^2 + 22 = 0.
Clearly, (4s - 13) < 0. Since s > 0, s != 1, s != 2, we must have s = 3. If these values are now substituted into eqs. 1, there results a system of three linear eqs. in three unknowns:
x + z + u = 2s + 1
2x + u = 4z + 4
2x + z = 6u + 9.
We easily find that x(=y) = 96, z = 57, u = 40.
III) The integers x, y, z, u are two pairs of equal numbers.
Assume that x = y and z = u. In this event, the first of the eqs. 1 yields x = 2z - 1; if this is substituted in (2), we obtain
x = (2t - 3)z + t^2 + 1, and so (2t - 5)z + t^2 + 2 = 0.
It follows that (2t - 5) < 0, and since t > 0, t != 1, we have t = 2. Eqs. 1 may now be written
x + 2z = 2x + 1
2x + 5 = 4z + 4
whence x(=y) = 11, z(=u) = 6.
IV) Three of the integers x, y, z, u are equal.
It is necessary to consider two cases:
a) If y = z = u, then eqs (3) and the first of eqs (1) take of the form
x = 3(2w - 3)y + 3w^2 + 1
2x = 3y - 1
and these, clearly, are inconsistent.
b) If x = y = z, then the first of eqs. (1) is 2x + u = 2x + 1, from which we find u = 1.
The last of eqs. 1 becomes
3x = 2su + s^2 = 2s + s^2
x = s(s + 2)/3.
But x must be an integer; hence either s or (s+2) must be divisible by 3. That is, s = 3k, x = k(3k+2), or s = (3k - 2), x = (3k - 2)k. Here k is an arbitrary integer.
V) All the numbers x, y, z, u are the same.
In this case, the first of eqs. 1 yields 3x = 2x + 1, x = 1.
================================
Hence we have the following solutions:
1) x = y = 96, z = 57, u = 40.
2) x = y = 11, z = u = 6.
3) x = y = z = k(3k +- 2), u = 1.
4) x = y = z = u = 1.
==================================
UFA!!!! Q.E.D.
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lijko
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Re: ALWAYS A PERFECT SQUARE
« Reply #12 on: Sep 14th, 2006, 5:52pm » |
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a similar problem was on the SAT math, but it was just pure logic.
it was what 4 perfect square numbers add up to equal 135 but when subtracted from 135 equal a perfect square. or something like that.
there was some sort of really short equation way but the answers were like 4 25 16 36 i think. It was about 3 years ago, SAT. If anyone wants to actually find the problem and correct me, by my guess
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