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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
10 months, 1 week 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 · 10 months, 1 week 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 · 10 months, 1 week 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 · 10 months, 1 week ago

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@Nihar Mahajan 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 · 10 months, 1 week ago

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We can use matrices to find the values of a,b,c Tejas Khairnar · 10 months, 1 week ago

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@Tejas Khairnar 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 · 10 months, 1 week 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 · 10 months, 1 week ago

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@Calvin Lin Sir, can you explain how you got this? I'm sorry I do not know this method. :) Aditya Kumar · 10 months, 1 week ago

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@Aditya Kumar 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 · 10 months, 1 week ago

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@Calvin Lin Wow! Nice way! Aditya Kumar · 10 months, 1 week ago

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@Calvin Lin Don't you mean "of the form \( 1 \pmod{4} \)"? Siddhartha Srivastava · 10 months, 1 week ago

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@Siddhartha Srivastava Oooops, yes I do. Edited. Thanks! Calvin Lin Staff · 10 months, 1 week 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 · 10 months, 1 week ago

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@Siddharth Singh It is wrong to simply assume that \(c=0\). We must show that the answer is independent of \(c\). Julian Poon · 10 months, 1 week ago

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Comment deleted 10 months ago

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@Aditya Kumar Yes ,this method can be used for JEE. But I don't think it will hold good for olympiads like RMO. Nihar Mahajan · 10 months, 1 week ago

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@Nihar Mahajan That's true but I've never seen such a question in rmo. Aditya Kumar · 10 months, 1 week ago

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@Aditya Kumar @Aditya kumar ,Interesting it really works.Is this method always applicable in such problems. Siddharth Singh · 10 months, 1 week ago

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@Siddharth Singh No! It isn't. This kind of solutions assumes that the final answer is not dependent on the variable that has been left out. To me this is the worst way to "solve" a question because it has assumptions embedded in it and is thus not a proper solution. That is like asking:

What's the value of \(1+a\)?

And then you assume that \(a = 1\) and say "hence the answer is \(2\)". Julian Poon · 10 months, 1 week ago

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@Julian Poon That's true. I didn't give it a deep thought to it. Can u suggest a way to solve similar but unfriendly questions? Aditya Kumar · 10 months, 1 week ago

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@Aditya Kumar There is no universal way to solve these questions, it just comes down to skill... Julian Poon · 10 months, 1 week ago

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@Julian Poon Ok. Thanks for clearing! I'll delete my solution now! Aditya Kumar · 10 months, 1 week ago

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@Aditya Kumar An apple a day , keeps the doctor away... It always clears our doubts :P Nihar Mahajan · 10 months, 1 week ago

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@Nihar Mahajan Haha...true that. Aditya Kumar · 10 months, 1 week ago

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