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

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

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

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

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TopNewestDenote 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 · 1 year, 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 · 1 year, 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 · 1 year, 4 months ago

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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 · 1 year, 4 months ago

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We can use matrices to find the values of a,b,c – Tejas Khairnar · 1 year, 4 months ago

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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 · 1 year, 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 · 1 year, 4 months agoLog in to reply

– Aditya Kumar · 1 year, 4 months ago

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

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 · 1 year, 4 months ago

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– Aditya Kumar · 1 year, 4 months ago

Wow! Nice way!Log in to reply

– Siddhartha Srivastava · 1 year, 4 months ago

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

– Calvin Lin Staff · 1 year, 4 months ago

Oooops, yes I do. Edited. Thanks!Log in to reply

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 · 1 year, 4 months ago

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– Julian Poon · 1 year, 4 months ago

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

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– Nihar Mahajan · 1 year, 4 months ago

Yes ,this method can be used for JEE. But I don't think it will hold good for olympiads like RMO.Log in to reply

– Aditya Kumar · 1 year, 4 months ago

That's true but I've never seen such a question in rmo.Log in to reply

@Aditya kumar ,Interesting it really works.Is this method always applicable in such problems. – Siddharth Singh · 1 year, 4 months ago

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

And then you assume that \(a = 1\) and say "hence the answer is \(2\)". – Julian Poon · 1 year, 4 months ago

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– Aditya Kumar · 1 year, 4 months ago

That's true. I didn't give it a deep thought to it. Can u suggest a way to solve similar but unfriendly questions?Log in to reply

– Julian Poon · 1 year, 4 months ago

There is no universal way to solve these questions, it just comes down to skill...Log in to reply

– Aditya Kumar · 1 year, 4 months ago

Ok. Thanks for clearing! I'll delete my solution now!Log in to reply

– Nihar Mahajan · 1 year, 4 months ago

An apple a day , keeps the doctor away... It always clears our doubts :PLog in to reply

– Aditya Kumar · 1 year, 4 months ago

Haha...true that.Log in to reply