Have you tried multivariable polynomials?

Algebra Level 5

P ( x + y , x − y ) = 2 P ( x , y ) P ( 1 , 1 ) = 4 P ( 2 , 3 ) = 23 \begin{aligned} P(x+y, x-y)&=&2 P(x,y) \\ P(1,1)&=&4 \\ P(2,3)&=&23 \end{aligned} There is just one non-zero polynomial P ( x , y ) P(x,y) with real coefficients that satisfies the properties given above.

What will be the value of P ( 8 7 , − 4 7 ) P\left (\dfrac{8}{\sqrt{7}},-\dfrac{4}{\sqrt{7}} \right) ?

This problem is from second stage of Iranian mathematical olympiads 2007 and belongs to this set .


The answer is 16.

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Kazem Sepehrinia by Kazem Sepehrinia , 31, Iran

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3 solutions

P ( x + y , x − y ) = 2 P ( x , y ) P(x+y,x-y)=2P(x,y)

⇒ P ( ( x + y ) + ( x − y ) , ( x + y ) − ( x − y ) ) = P ( 2 x , 2 y ) = 2 P ( x + y , x − y ) = 4 P ( x , y ) \Rightarrow P((x+y)+(x-y),(x+y)-(x-y))=P(2x,2y)=2P(x+y,x-y)=4P(x,y)

⇒ P ( 2 x , 2 y ) = 4 P ( x , y ) \Rightarrow P(2x,2y)=4P(x,y)

This means that our multivariate polynomial is second degree and is of the form:

P ( x , y ) = a x 2 + b y 2 + c x y P(x,y)=a{ x }^{ 2 }+b{ y }^{ 2 }+cxy

Straight away we have:

P ( 1 , 1 ) = a + b + c = 4 P(1,1)=a+b+c=4 , P ( 2 , 3 ) = 4 a + 9 b + 6 c = 23 P(2,3)=4a+9b+6c=23 . Also, using the given identity:

P ( 1 , 1 ) = P ( 1 + 0 , 1 − 0 ) = 2 P ( 1 , 0 ) , ⇒ P ( 1 , 0 ) = a = 2 P(1,1)=P(1+0,1-0)=2P(1,0), \Rightarrow P(1,0)=a=2

Immediately: P ( x , y ) = 2 x 2 + y 2 + x y P(x,y)=2{ x }^{ 2 }+{ y }^{ 2 }+xy

Hence: P ( 8 7 , − 4 7 ) = 16 P\left (\frac{8}{\sqrt{7}},-\frac{4}{\sqrt{7}} \right)= 16

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Nicely done!Much easier than my own solution

Arian Tashakkor - 6 years, 1 month ago

Thanks for the point on second degree. I was trying to prove it.

Chew-Seong Cheong - 6 years, 1 month ago
Arian Tashakkor
May 5, 2015

I'm not quite sure if you picked those numbers on purpose or anything but the numbers you gave as the info. makes the question pretty easy to solve!

P ( x + y , x − y ) = 2 P ( x , y ) → P(x+y, x-y)=2 P(x,y) \rightarrow

If: y = 0 ⇒ P ( x , x ) = 2 P ( x , 0 ) y=0 \Rightarrow P(x,x)=2P(x,0)

If: y = x ⇒ P ( 2 x , 0 ) = 2 P ( x , x ) y=x \Rightarrow P(2x,0)=2P(x,x)

Now we have P ( 1 , 1 ) = 4 P(1,1)=4

And also that, P ( 2 , 2 ) = 2 P ( 2 , 0 ) = 4 P ( 1 , 1 ) P(2,2)=2P(2,0)=4P(1,1)

Hence, P ( 2 , 2 ) = 4 × 4 = 16 P(2,2)=4 \times 4 =16

Sorry if I disappointed you! xD

P.S:In fact there's no use for the second piece of information and knowing P ( 1 , 1 ) = 4 P(1,1)=4 suffices.

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Not really sure about this solution but here's the 2nd solution for the edited question:( BRACE YOURSELVES!WALL OF TEXT INCOMING!!! )

#2 Solution:

P ( x , y ) = 2 P ( x + y 2 , x − y 2 ) = 4 P ( x + y 2 + x − y 2 2 , x + y 2 − x − y 2 2 ) = 4 P ( x 2 , y 2 ) P(x,y)=2P(\frac{x+y}{2} , \frac {x-y}{2} ) = 4P(\frac {\frac {x+y}{2} +\frac {x-y}{2}}{2} , \frac {\frac {x+y}{2} -\frac {x-y}{2}}{2}) = 4P(\frac{x}{2} , \frac {y}{2})

Now suppose the coefficient for the term x a y b x^a y^b is K,hence the coefficient for the same term in 4 P ( x 2 , y 2 ) 4P(\frac{x}{2} , \frac {y}{2}) is 4 K × 2 − a − b 4K \times 2^{-a-b} (You can check it by hand if you don't get the 2 a − b 2^{a-b} part!),thus, we have:

K = 4 K × 2 − a − b → 4 × 2 − a − b = 1 ⇒ a + b = 2 ; ∀ ( a , b ) I K=4K \times 2^{-a-b} \rightarrow 4 \times 2^{-a-b}=1 \Rightarrow a+b=2 \quad ;\quad \forall (a,b) \quad \quad \mathbb{I}

I \mathbb{I} gives us all we need to solve the problem!The fact that a + b = 2 ; ∀ ( a , b ) a+b=2 \quad ; \quad \forall (a,b) means that the mentioned polynomial consists of terms x 2 , y 2 , x y x^2 , y^2 , xy only!

Now let's assume P ( x , y ) = a x 2 + b x y + c y 2 P(x,y)=ax^2+bxy+cy^2 using the information from the problem we get:

2P(x,y)=2ax^2+2bxy+2cy^2=P(x+y,x-y)=\text {(#after simplifying#)}

( a + b + c ) x 2 + 2 ( a − c ) x y + ( a + c − b ) y 2 (a+b+c)x^2+2(a-c)xy+ (a+c-b)y^2

Hence:

a + b + c = 2 a , 2 ( a − c ) = 2 b , a + c − b = 2 c a+b+c=2a , 2(a-c)=2b , a+c-b=2c from all of which we can deduce a = b + c a=b+c

Hence:

P ( x , y ) : ( b + c ) x 2 + b x y + c y 2 P(x,y): (b+c)x^2+bxy+cy^2

But wait there's still much more to do!The result we got from the last line indicates that there's an infinite number of polynomials only considering P ( x + y , x − y ) = 2 P ( x , y ) P(x+y, x-y)=2 P(x,y) now with the other 2 pieces of info. hopefully we can tame this wild problem!

P ( 1 , 1 ) = 4 → b + c + b + c = 2 b + 2 c = 4 → b + c = 2 I I P(1,1)=4 \rightarrow b+c+b+c=2b+2c=4 \rightarrow b+c=2 \mathbb{II}

P ( 2 , 3 ) = 23 → 2 × 4 + 6 b + 9 c = 8 + 6 b + 6 c + 3 c = 23 → 3 c = 3 P(2,3)=23 \rightarrow 2 \times 4 + 6b +9c =8+6b+6c+3c=23 \rightarrow 3c=3

→ c = 1 → using II → b = 1 \rightarrow c=1 \rightarrow \text {using II } \rightarrow b=1

⇒ P ( x , y ) = 2 x 2 + x y + y 2 → P ( 8 7 , − 4 7 ) = 112 7 = 16 \Rightarrow P(x,y)=2x^2+xy+y^2 \rightarrow P(\frac{8}{\sqrt{7}},-\frac{4}{\sqrt{7}}) = \frac {112}{7} = 16

EDIT:Sorry about the wrong answer.I edited my answer and the answer is now 16.(It was almost 3 in the morning when I posted the solution so you know ... xD)

Arian Tashakkor - 6 years, 1 month ago

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Very good Arian, nice job!

Kazem Sepehrinia - 6 years, 1 month ago

Thanks!Nice problem by the way.Could you post a functional equation problem as well?I really enjoy solving them.

Arian Tashakkor - 6 years, 1 month ago

OMG, I made a silly mistake Arian, I must change the question.

Hey Arian, I wanna be in contact with you, Can I have your gmail or something?

Kazem Sepehrinia - 6 years, 1 month ago

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Yes of course you should!And by the way rethink that "there's only one non-zero polynomial" part.Are you sure about that?Cuz, I got different results.In fact there is an infinite number of polynomials, all of which can be defined into just "ONE" using the information you provided. #RIPenglish :/

Arian Tashakkor - 6 years, 1 month ago

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I've edited the question, see and aware me if there is a false anywhere.

Kazem Sepehrinia - 6 years, 1 month ago

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@Kazem Sepehrinia – Yup that seems pretty well.Now the process of solving shall begin!Long road ahead of me I guess xD

As of your question about my email of course!

Email: [email protected] I'd be glad to be able to speak to some fellow math-freak geniuses at anytime !!! (btw , nice job entering "Sharif univ." )

Arian Tashakkor - 6 years, 1 month ago

From P ( x + y , x − y ) = 2 P ( x , y ) \space P(x+y,x-y) = 2P(x,y)

\(\begin{array} {} \Rightarrow & P(1,1) = 4 \\ & P(2,0) = P(1+1,1-0) = 2P(1,1) = 8 \\ & P(2,2) = 16 \\ & P(4,0) = 32 \\ & P(4,4) = 64 \\ & P(8,0) = 128 \\ & ... \end{array} \)

From the above, it can be seen that P ( x , y ) = 2 x 2 + Q ( x , y ) \space P(x,y) = 2x^2 + Q(x,y)\space and there is no constant term. Assuming Q ( x , y ) = a x y + b y 2 \space Q(x,y) = axy + by^2 \space and using the following cases, we have:

{ P ( 2 , 2 ) = 8 + 4 a + 4 b = 16 ⇒ a + b = 2 . . . ( 1 ) P ( 2 , 3 ) = 8 + 6 a + 9 b = 23 ⇒ 2 a + 3 b = 5 . . . ( 2 ) \begin{cases} P(2,2) = 8 + 4a + 4b = 16 & \Rightarrow a+b = 2 &...(1) \\ P(2,3) = 8 + 6a + 9b = 23 & \Rightarrow 2a+3b = 5 &...(2) \end{cases}

⇒ Eq.2 − 2 × Eq.1 : b = 1 ⇒ a = 1 ⇒ P ( x , y ) = 2 x 2 + x y + y 2 ⇒ P ( 8 7 , − 4 7 ) = 128 7 − 32 7 + 16 7 = 16 \Rightarrow \text{Eq.2} - 2\times \text{Eq.1}: \quad b = 1 \quad \Rightarrow a = 1 \\ \Rightarrow P(x,y) = 2x^2+xy+y^2 \\ \Rightarrow P\left(\frac{8}{\sqrt{7} }, -\frac{4}{\sqrt{7}}\right) = \dfrac{128}{7}-\dfrac{32}{7} + \dfrac{16}{7} = \boxed{16}

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