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        riddles >> hard >> ALWAYS A PERFECT SQUARE
        
(Message started by: pcbouhid on Nov 29^th, 2005, 8:23am)

Title: ALWAYS A PERFECT SQUARE
Post by pcbouhid on Nov 29^th, 2005, 8:23am

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.

Title: Re: ALWAYS A PERFECT SQUARE
Post by Barukh on Nov 29^th, 2005, 8:52am

Paolo, the trivial answer would be [hide]1, 1, 1, 1[/hide], but that's probably not what you want. ;)

Title: Re: ALWAYS A PERFECT SQUARE
Post by JocK on Nov 29^th, 2005, 10:22am

Extending Barukh's solution, an infinite number of (slightly) less trivial answers is given by:

[hide]
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, ...

[/hide]

Title: Re: ALWAYS A PERFECT SQUARE
Post by JohanC on Nov 29^th, 2005, 1:02pm

Two other solutions, not including the number 1:
[hide]6 6 11 11
40 57 96 96[/hide]

Title: Re: ALWAYS A PERFECT SQUARE
Post by JocK on Nov 29^th, 2005, 1:44pm

Would there be a solution with four distinct natural numbers?

Title: Re: ALWAYS A PERFECT SQUARE
Post by SMQ on Nov 29^th, 2005, 2:05pm

Not with the largest number < [hide]2000, which is as far as my exhaustive computerized search has reached so far... ;)[/hide]

[edit][hide]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...[/hide][/edit]

--SMQ

Title: Re: ALWAYS A PERFECT SQUARE
Post by pcbouhid on Nov 30^th, 2005, 10:16am

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: [hide]this case is not possible.[/hide]

2) Precisely two of the integers are equal: [hide]x = y = 96, z = 57, u = 40.[/hide]

3) The numbers are two pairs of equal numbers:[hide] x = y = 11, z = u = 6.[/hide]

4) Three of the numbers are equal: [hide]u = 1, and x = y = z = k(3k+2) or k(3k-2), k an arbitrary integer.[/hide]

5) All the numbers are the same: [hide]x = y = z = u = 1. [/hide]

Title: Re: ALWAYS A PERFECT SQUARE
Post by JocK on Nov 30^th, 2005, 10:43am

Do you mean to say that we found all integer solutions..?

Title: Re: ALWAYS A PERFECT SQUARE
Post by pcbouhid on Dec 1^st, 2005, 6:02am

Exactly!!!!

Title: Re: ALWAYS A PERFECT SQUARE
Post by JocK on Dec 1^st, 2005, 9:25am

Can you prove that?

(Feels good... being given the opportunity for bouncing back a question .. ;D )

Title: Re: ALWAYS A PERFECT SQUARE
Post by pcbouhid on Dec 2^nd, 2005, 5:51am

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.

Title: Re: ALWAYS A PERFECT SQUARE
Post by pcbouhid on Dec 2^nd, 2005, 9:28am

Detailed solution:

[hide]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.[/hide]
==================================

UFA!!!! Q.E.D.

Title: Re: ALWAYS A PERFECT SQUARE
Post by lijko on Sep 14^th, 2006, 5:52pm

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