Maths problem

[Sosaria]

Calling the maths experts:

 

Given three unknown square numbers add up to 109, what are these numbers?

 

The conditions are that these numbers are all not equal, and they are whole numbers.

 

Is there a systematic, methodical way to determine these numbers, without resorting to "guess and check"?

[1fast1]

Calling the maths experts:

 

Given three unknown square numbers add up to 109, what are these numbers?

 

The conditions are that these numbers are all not equal, and they are whole numbers.

 

Is there a systematic, methodical way to determine these numbers, without resorting to "guess and check"?

 

A solution set (3,6,8) is quickly found by observing that 109 = 100 + 9 = 10^2 + 3^2.

 

10^2 = 5^2.2^2, and since (3,4,5) is a well known primitive Pythagorean triple, we can quickly deduce that the other numbers are 2*3 = 6 and 2*4 = 8.

 

Proving uniqueness is more challenging, and I'm still working on it.

 

If you want a more systematic method than this, I wager it'll be difficult. This is called a quadratic Diophantine equation, and the theory here is quite deep, involving Pell-like equations and what-not. Quite cheem.

[Sosaria]

Thanks! I think that's already systematic enough.

[Mcf777]

which level is this type of Math?

 

So cheem ka logi

I never learn it before during my Osss level.

[Jamesc]

Well solved by TF4

 

Edited July 26, 2013 by Jamesc

[1fast1]

OK, I've found a way to prove uniqueness of the result for the squares of three natural numbers a, b and c. Not the most elegant, but it'll do. It also allows a slightly more systematic solution of the equation by elementary means.

 

The equation can be represented by a^2 + b^2 + c^2 = 109. The RHS (right hand side) is odd (it's also prime, but that's not that relevant here).

 

Clearly, a, b and c can't all be even. If exactly two of them are odd, they will sum to an even number, which when added to the last even number, will give an even number (which clearly can't be 109). Therefore at least one of a,b and c must be odd. That admits of 2 possibilities, either the LHS is odd + odd + odd or odd + even + even.

 

Let's eliminate one of these two possibilities. Observe that 109 is of the form 4k + 1 where k is an integer (simply put, this means one more than a multiple of four).

 

It can quite easily be proven that a perfect square can only be of the form 4s (if even) or 4s+1 (if odd), where s is an integer. Since we know that at least one of a^2, b^2 and c^2 is odd, let's subtract that from the RHS. With no loss of generality , we can say that a^2 is odd.

 

So we have: b^2 + c^2 = 109-a^2, where a^2 is odd (implying a is odd). Since a has to be of the form 4s + 1, and 109 = 4k + 1, we get:

 

b^2 + c^2 = 4(k - s)

 

So b^2 + c^2 is a multiple of 4. Clearly b and c are both even, and we can now express b = 2m, c = 2n, giving:

 

(2m)^2 + (2n)^2 = 4(k-s)

 

Hence,

 

m^2 + n^2 = k - s

 

Recall that k = 27 because 109 = 4*27 + 1. So we're left with the much simpler equation:

 

m^2 + n^2 = 27 - s

 

Can we go eliminate further possibilities? Yes we can. Remember that we originally defined s by c^2 = 4s + 1. Let's rearrange that:

 

c^2 - 1 = 4s

 

(c+1)(c-1) = 4s

 

s = (1/4)*(c+1)(c-1)

 

Clearly, s has to be a natural number. The only way that can happen is for the product (c+1)(c-1) to be divisible by 4, which implies that both factors are even. So c - 1 = 2u, c + 1 = 2(u+1) giving s = u*(u+1). Therefore, s has to be the productive of consecutive natural numbers!

 

That makes things really simple when we do the final elimination. We now have to find natural numbers m and n such that m^2 + n^2 = 27 - s, where s is itself the product of two consecutive natural numbers.

 

The only values of s that can be admitted are clearly: 1*2, 2*3, 3*4 and 4*5, since anything higher gives a negative value on the RHS.

 

So we have to consider solutions to:

 

m^2 + n^2 = 25

 

m^2 + n^2 = 21

 

m^2 + n^2 = 15

 

m^2 + n^2 = 7

 

 

The second to fourth equations can be solved by subtracting all possible values of n^2 from the RHS, e.g. for the second equation:

 

n^2 can be 1, 4, 6, 9, 16 but no higher. It's quickly seen that none of these values can give a valid m^2.

 

Ditto for equations 3 (test n^2 = 1, 4, 9) and 4 (test only n^2 = 1, 4).

 

The first equation, when similarly exhaustively tested (for n^2 = 1, 4, 9, 16), yields a valid solution for m^2 only for n^2 = 9 (or n = 3) and n^2 = 16 (or n = 4). In this case, m^2 = 25-16 = 9 or 25-9 = 16 respectively, so either (m = 3, n=4) or (m=4, n=3). These are basically interchangeable solutions.

 

Therefore m = 3, n = 4 and s = 2 OR m = 4, n = 3, s = 2 gives the only valid solution sets, which corresponds to a^2 = 4*2 + 1 = 9 (=3^2), b^2 = 4*3^2 = 36 (=6^2) and c^2 = 4*4^2 = 64 (=8^2) or a^2 = 9, b^2 = 64, c^2 = 36. As mentioned, these two are essentially redundant and interchangeable.

 

Thus (3,6,8) is proven to be the only valid unique solution set for natural a, b and c. (QED)

 

[Note: Technically, zero is a whole number (integer), so (3,0,10) (keeping the order as above) is also a valid solution to the original problem if there is no requirement for a, b and c to all be natural numbers (which excludes zero). This solution occurs when you consider the equation m^2 + n^2 = 25 and allow for n^2 = 25, which gives m^2 = 0. So if this is allowed, there are, in fact, two unique solutions in the integers, namely: (3,6,8) and (0,3,10) when the solutions are rearranged in ascending order].

Edited July 26, 2013 by Turboflat4

[Sk65]

A solution set (3,6,8) is quickly found by observing that 109 = 100 + 9 = 10^2 + 3^2.

 

10^2 = 5^2.2^2, and since (3,4,5) is a well known primitive Pythagorean triple, we can quickly deduce that the other numbers are 2*3 = 6 and 2*4 = 8.

 

Proving uniqueness is more challenging, and I'm still working on it.

 

If you want a more systematic method than this, I wager it'll be difficult. This is called a quadratic Diophantine equation, and the theory here is quite deep, involving Pell-like equations and what-not. Quite cheem.

 

knn,

I study so many years, also never see anything more cheem than what you have just typed....

 

heng after Air level no need to do math already

Edited July 26, 2013 by Sk65

[Voodooman]

 

 

You won book prizes and you can solve super cheem math question.

 

I have always worshipped the tablewiper (wipe tables also can drive BMW and stay in big bungalow, better than me, everyday wipe kachen, ie. cleaning up mess created by others, and still money not enough), I think I have found a new idol.

 

This is impressive, I am now very curious to know what you do for a living.

 

 

[VenomMX]

Never thought i would see "wlog" in this forum..

[Coltplussport]

Calling the maths experts:

 

Given three unknown square numbers add up to 109, what are these numbers?

 

The conditions are that these numbers are all not equal, and they are whole numbers.

 

Is there a systematic, methodical way to determine these numbers, without resorting to "guess and check"?

 

Very easy,

 

1 sq =1

2 sq =4

3 sq =9

4 sq =16

5 sq =25

6 sq =36

7 sq =49

8 sq =64

9 sq =81

10 sq =100

 

So it can't be 1, cause the other 2 cannot be 108.

 

It can't be 2, because the other 2 cannot be 105.

 

It is 3 because the other 2 is 6 and 8. 3sq is 9+ 6 sq is 36 +8 sq is 64.

 

 

 

 

 

 

So the only answer got to be

[Hydrocarbon]

OK, I've found a way to prove uniqueness of the result for the squares of three natural numbers a, b and c. Not the most elegant, but it'll do. It also allows a slightly more systematic solution of the equation by elementary means.

 

Woah, this is so cheem leh.. Thought I'd see something do to with algebra, but when I see your reply, my eyes go already..

[RadX]

if got nuclear bomb in sg, i know who.....

 

that is why, in mcf, dun play play, we have all the expertise here, and well enough, to share on such like these

 

thanks dey! [thumbsup]

[Wind30]

actually by eyeball, you will see 100+9=109

 

The question is actually asking if 0 is a whole number. If it is, the answer is very simple.

[Porker]

Very easy,

 

1 sq =1

2 sq =4

3 sq =9

4 sq =16

5 sq =25

6 sq =36

7 sq =49

8 sq =64

9 sq =81

10 sq =100

 

So it can't be 1, cause the other 2 cannot be 108.

 

It can't be 2, because the other 2 cannot be 105.

 

It is 3 because the other 2 is 6 and 8. 3sq is 9+ 6 sq is 36 +8 sq is 64.

 

 

 

 

 

 

So the only answer got to be

 

Short, simple and elegant. Best post here

[Porker]

Calling the maths experts:

 

Given three unknown square numbers add up to 109, what are these numbers?

 

The conditions are that these numbers are all not equal, and they are whole numbers.

 

Is there a systematic, methodical way to determine these numbers, without resorting to "guess and check"?

 

This sounds very GMAT

[Rayleigh]

Very easy,

 

1 sq =1

2 sq =4

3 sq =9

4 sq =16

5 sq =25

6 sq =36

7 sq =49

8 sq =64

9 sq =81

10 sq =100

 

So it can't be 1, cause the other 2 cannot be 108.

 

It can't be 2, because the other 2 cannot be 105.

 

It is 3 because the other 2 is 6 and 8. 3sq is 9+ 6 sq is 36 +8 sq is 64.

 

 

 

 

 

 

So the only answer got to be

 

Bro, criteria no guess and check liao. Answer correct but working wrong. No marks awarded.

[1fast1]

Very easy,

 

1 sq =1

2 sq =4

3 sq =9

4 sq =16

5 sq =25

6 sq =36

7 sq =49

8 sq =64

9 sq =81

10 sq =100

 

So it can't be 1, cause the other 2 cannot be 108.

 

It can't be 2, because the other 2 cannot be 105.

 

It is 3 because the other 2 is 6 and 8. 3sq is 9+ 6 sq is 36 +8 sq is 64.

 

 

 

 

 

 

So the only answer got to be

 

I'm sorry - but I have to say that it's not clear what you're doing here.

 

Why can't the sum of the other 2 be equal to 108 or 105?

 

You're correct that the max value of any of three can only be 10. But even if so, the sum of squares of two numbers both less than ten can still add up to numbers like 106 (5^2 + 9^2), so how exactly have you eliminated these possibilities?

 

Basically if you're using completely naive "guess and check". you actually have to subtract perfect squares from 109, then test which of those are expressible as the sum of two squares themselves. This is tedious, and I believe the OP does not want it done this way.

 

But if you had another insight that allows you to "automatically" exclude the possibility that the sum of the other two squares can be those numbers, I'm curious to know. Always looking for a more elegant method.

Edited July 26, 2013 by Turboflat4

[1fast1]

You won book prizes and you can solve super cheem math question.

 

I have always worshipped the tablewiper (wipe tables also can drive BMW and stay in big bungalow, better than me, everyday wipe kachen, ie. cleaning up mess created by others, and still money not enough), I think I have found a new idol.

 

This is impressive, I am now very curious to know what you do for a living.

 

Please don't idolise me lah bro. I'm just a humble yisheng, and an especially humble one at that - definitely no surgeon earning the big bucks.

 

Continue to idolise T2, after all, I do too!

 

I've always been keen on Math/Physics - had a PSC OMS scholarship to do EE at Caltech, but turned it down to do Medicine here so that I obey father's wishes and be close to home. Bit late to regret now.