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Algebra Level 5

How many ordered quadruples of non-negative real numbers are there that satisfies:

$\begin{aligned} a + b + c + d = 4 \\ a^2bc + b^2cd + c^2da + d^2 ab = 4 \\ \end{aligned}$

Now prove this .

1 9 5 Infinitely many 13

1 solution

Calvin Lin Staff
May 20, 2015

[This is not a complete solution. See Nihar's comment below]

There are only 5 solutions, namely $(1, 1, 1, 1), (2, 1, 1, 0), (0, 2, 1, 1), (1, 0, 2, 1), (1, 1, 0, 2)$ .

Let me post my approach and tell me whether it is wrong or incomplete.We equate the equations :

$a+b+c+d = a^2bc+b^2cd+c^2da+d^2ab \\ \Rightarrow a-a^2bc+b-b^2cd+c-c^2da+d-d^2ab =0\\ \Rightarrow a(1-abc)+b(1-bcd)+c(1-cda)+d(1-dab)=0$

At glance we get the trivial solution $a,b,c,d=0$ which is not possible.So for the expression to be $0$ ,

$abc=bcd=cda=dab=1$

We multiply all these equations to get :

$(abcd)^3=1 \Rightarrow abcd=1$

By $AM - GM$ for positive real numbers which satisfy $abcd=1$ and $a+b+c+d=4$ , we get only one solution $a=b=c=d=1$ .

Edit Credits : Pi Han Goh.

Nihar Mahajan - 6 years ago

Great work!

Sravanth C. - 6 years ago

Funny , I was so close ... !Well done Nihar.

Arian Tashakkor - 6 years ago

Just because $abcd=1$ and $a+b+c+d=4$ , how do you get infinite number of non-negative solution?

In fact, you should only get one solution by AM GM.

Pi Han Goh - 6 years ago

I think my solution is incomplete.I have left out many cases.So shall I delete it ?

Nihar Mahajan - 6 years ago

@Nihar Mahajan – Just remove the last line. and change it to "if they are all positive, then $a=b=c=d=1$ is the only solution by AM-GM."

You can check the formal solution in the report section. =D

Pi Han Goh - 6 years ago

@Pi Han Goh – Ok , Thanks!

Nihar Mahajan - 6 years ago

How to get that quadruple?

Pranjal Jain - 6 years ago

@Calvin Lin can we imagine the given expression as a level surface of the 4 d curve ? ....... and hence there are infinitely many solutions ? ...... please correct me if im wrong.

Abhinav Raichur - 6 years ago

If $a = 1$ , $b = k$ , $c = 2 - k$ , and $d = 1$ , then

$\begin{aligned} &a^2 bc + b^2 cd + c^2 da + d^2 ab \\ &= k(2 - k) + k^2 (2 - k) + (2 - k)^2 + k \\ &= -k^3 + 2k^2 - k + 4. \end{aligned}$

Jon Haussmann - 6 years ago

Thanks. I was thinking of different versions of the problem and got confused. I have updated the answer to 5.

Calvin Lin Staff - 6 years ago

same observation here

Abhinav Raichur - 6 years ago

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Edit Credits : Pi Han Goh.

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