An algebra question from pre RMO 2015

Let \(a,b,c\) be reals such that \(a-7b+8c=4\) and \(8a+4b-c=7\)

Then find the value of \(a^2-b^2+c^2\)

Note by Nihar Mahajan
3 years, 5 months ago

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

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

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

So \(a^2-b^2+c^2=(a+b)(a-b)+c^2=1-c^2+c^2=1\).

Xuming Liang - 3 years, 4 months ago

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Nice question

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

Calvin Lin Staff - 3 years, 5 months ago

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

Siddhartha Srivastava - 3 years, 5 months ago

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Oooops, yes I do. Edited. Thanks!

Calvin Lin Staff - 3 years, 5 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, 4 months ago

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Direct Hint: \( 65 = 1^2 + 8^2 = 7^2 + 4^2 \)

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

Calvin Lin Staff - 3 years, 4 months ago

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@Calvin Lin Wow! Nice way!

Aditya Kumar - 3 years, 4 months ago

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

Tejas Khairnar - 3 years, 4 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 \( a^2 - b^2 + c^2 \) 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, 4 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 \(b\) from both the equations,

\[b=\dfrac{a+8c-4}{7}=\dfrac{7+c-8a}{4}\]

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

This solution is incomplete since it assumes that \(13a,13c\) are integers.

Nihar Mahajan - 3 years, 4 months ago

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

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

Calvin Lin Staff - 3 years, 4 months ago

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Slightly easier approach that follows from my hint. Consider:

\[ (a-7b+8c)^2 + (8a+4b-c)^2 \]

Notice that the cross terms cancel out.


This exploits the symmetry of the values, and the observation that \( 1^2 + 8^2 = 4^2 + 7^2\).

Calvin Lin Staff - 3 years, 4 months ago

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2^{0}

Saroja Gudimetla - 8 months, 4 weeks ago

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0

Kuhu Raychaudhuri - 8 months, 3 weeks ago

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Eliminating \(a,b,c\) one by one from both equations we get 3 equations:

\(13a+5b=12\) ...1

\(13c-12b=5\) ...2

\(12a-5c=13\) ...3

Again by elimination of each of \(a,b\) and \(c\),we get \(a=\frac{13}{12},b=-\frac{5}{12},c=0\)

Putting them in \(a^{2}-b^{2}+c^{2}\) we get \(\boxed{1}\).

Siddharth Singh - 3 years, 5 months ago

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

Julian Poon - 3 years, 5 months ago

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