An algebra question from pre RMO 2015

Let a,b,ca,b,c be reals such that a7b+8c=4a-7b+8c=4 and 8a+4bc=78a+4b-c=7

Then find the value of a2b2+c2a^2-b^2+c^2

Note by Nihar Mahajan
3 years, 11 months ago

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Denote the two equations (1)(1) and (2)(2) respectively:

(1)2(2)(1)-2(2) gives us a+b=23(1+c)a+b=\frac {2}{3}(1+c)

2(1)+(2)2(1)+(2) gives us ab=32(1c)a-b=\frac {3}{2}(1-c)

So a2b2+c2=(a+b)(ab)+c2=1c2+c2=1a^2-b^2+c^2=(a+b)(a-b)+c^2=1-c^2+c^2=1.

Xuming Liang - 3 years, 11 months ago

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Nice question

Cryptic Hint: 65=5×13 65 = 5 \times 13 . These are both primes of the form 1(mod4) 1 \pmod{4} .

Calvin Lin Staff - 3 years, 11 months ago

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Don't you mean "of the form 1(mod4) 1 \pmod{4} "?

Siddhartha Srivastava - 3 years, 11 months ago

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Oooops, yes I do. Edited. Thanks!

Calvin Lin Staff - 3 years, 11 months ago

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Sir, can you explain how you got this? I'm sorry I do not know this method. :)

Aditya Kumar - 3 years, 11 months ago

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Direct Hint: 65=12+82=72+42 65 = 1^2 + 8^2 = 7^2 + 4^2

Explanation of cryptic hint: By Fermat's Two square theorem, since 65=5×13=(22+12)(22+32) 65 = 5 \times 13 = ( 2^2 + 1^2 ) ( 2^2 + 3^2 ) , we can write 65 as the sum of two (positive) squares in two distinct ways.

Calvin Lin Staff - 3 years, 11 months ago

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@Calvin Lin Wow! Nice way!

Aditya Kumar - 3 years, 11 months ago

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We can use matrices to find the values of a,b,c

Tejas Khairnar - 3 years, 11 months ago

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You have 3 variables and 2 equations, so you do not have a completely determined system.

Though, the claim is that a2b2+c2 a^2 - b^2 + c^2 is independent of the actual values. The point of the question is to prove that the expression is always equal to 1.

Calvin Lin Staff - 3 years, 11 months ago

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I don't know whether this method is valid , but we learn from mistakes and hence, I try:

Computing for bb from both the equations,

b=a+8c47=7+c8a4b=\dfrac{a+8c-4}{7}=\dfrac{7+c-8a}{4}

Now simplifying the two equations at the right , we get 12a+5c=1312(13a)+5(13c)=13212a+5c=13\Rightarrow 12(13a)+5(13c)=13^2 which imitates the Pythagorean triplet 122+52=13212^2+5^2=13^2 and then comparing them we have 13a=12 , 13c=513a=12 \ , \ 13c=5 which when substituted in the either of the first equations , we have b=0b=0. Now substituting the acquired values in required expression we have a2b2+c2=1a^2-b^2+c^2=1.

This solution is incomplete since it assumes that 13a,13c13a,13c are integers.

Nihar Mahajan - 3 years, 11 months ago

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This method doesn't constitute a proof, since you're essentially substituting b=0,a=1213,c=513 b =0 , a = \frac{12}{13}, c = \frac{5}{13} .

If you want a proof by substitution, you could
1. Determine aa and bb in terms of cc only.
2. Substitute these values into a2b2+c2 a^2 - b^2 + c^ 2.
3. Hope that everything cancels out (which it should, if your arithmetic is correct).

Calvin Lin Staff - 3 years, 11 months ago

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Slightly easier approach that follows from my hint. Consider:

(a7b+8c)2+(8a+4bc)2 (a-7b+8c)^2 + (8a+4b-c)^2

Notice that the cross terms cancel out.


This exploits the symmetry of the values, and the observation that 12+82=42+72 1^2 + 8^2 = 4^2 + 7^2.

Calvin Lin Staff - 3 years, 11 months ago

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2^{0}

Saroja Gudimetla - 1 year, 2 months ago

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0

Kuhu Raychaudhuri - 1 year, 2 months ago

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Eliminating a,b,ca,b,c one by one from both equations we get 3 equations:

13a+5b=1213a+5b=12 ...1

13c12b=513c-12b=5 ...2

12a5c=1312a-5c=13 ...3

Again by elimination of each of a,ba,b and cc,we get a=1312,b=512,c=0a=\frac{13}{12},b=-\frac{5}{12},c=0

Putting them in a2b2+c2a^{2}-b^{2}+c^{2} we get 1\boxed{1}.

Siddharth Singh - 3 years, 11 months ago

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It is wrong to simply assume that c=0c=0. We must show that the answer is independent of cc.

Julian Poon - 3 years, 11 months ago

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