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Topic: ALWAYS A PERFECT SQUARE (Read 1540 times) 

pcbouhid
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ALWAYS A PERFECT SQUARE
« on: Nov 29^{th}, 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 29^{th}, 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 29^{th}, 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(3n2) : k = 1, 8, 21, 40, ... or: k = n(3n+2) : k = 5, 16, 33, 56, ...

« Last Edit: Nov 29^{th}, 2005, 11:59am by JocK » 
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x^{y}  y = x^{5}  y^{4}  y^{3} = 20; x>0, y>0.



JohanC
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Re: ALWAYS A PERFECT SQUARE
« Reply #3 on: Nov 29^{th}, 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 29^{th}, 2005, 1:44pm » 
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Would there be a solution with four distinct natural numbers?


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x^{y}  y = x^{5}  y^{4}  y^{3} = 20; x>0, y>0.



SMQ
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Re: ALWAYS A PERFECT SQUARE
« Reply #5 on: Nov 29^{th}, 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

« Last Edit: Nov 30^{th}, 2005, 5:39am by SMQ » 
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pcbouhid
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Re: ALWAYS A PERFECT SQUARE
« Reply #6 on: Nov 30^{th}, 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(3k2), 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 30^{th}, 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.
x^{y}  y = x^{5}  y^{4}  y^{3} = 20; x>0, y>0.



pcbouhid
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Re: ALWAYS A PERFECT SQUARE
« Reply #8 on: Dec 1^{st}, 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 1^{st}, 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.
x^{y}  y = x^{5}  y^{4}  y^{3} = 20; x>0, y>0.



pcbouhid
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Re: ALWAYS A PERFECT SQUARE
« Reply #10 on: Dec 2^{nd}, 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 2^{nd}, 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 14^{th}, 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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