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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
1 year, 8 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 $$a^2-b^2+c^2=(a+b)(a-b)+c^2=1-c^2+c^2=1$$. · 1 year, 8 months ago

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 $$1^2 + 8^2 = 4^2 + 7^2$$. Staff · 1 year, 8 months ago

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. · 1 year, 8 months ago

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). Staff · 1 year, 8 months ago

We can use matrices to find the values of a,b,c · 1 year, 8 months ago

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. Staff · 1 year, 8 months ago

Nice question

Cryptic Hint: $$65 = 5 \times 13$$. These are both primes of the form $$1 \pmod{4}$$. Staff · 1 year, 8 months ago

Sir, can you explain how you got this? I'm sorry I do not know this method. :) · 1 year, 8 months ago

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. Staff · 1 year, 8 months ago

Wow! Nice way! · 1 year, 8 months ago

Don't you mean "of the form $$1 \pmod{4}$$"? · 1 year, 8 months ago

Oooops, yes I do. Edited. Thanks! Staff · 1 year, 8 months ago

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 $$\boxed{1}$$. · 1 year, 8 months ago

It is wrong to simply assume that $$c=0$$. We must show that the answer is independent of $$c$$. · 1 year, 8 months ago

Comment deleted Nov 20, 2015

Yes ,this method can be used for JEE. But I don't think it will hold good for olympiads like RMO. · 1 year, 8 months ago

That's true but I've never seen such a question in rmo. · 1 year, 8 months ago

@Aditya kumar ,Interesting it really works.Is this method always applicable in such problems. · 1 year, 8 months ago

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 $$2$$". · 1 year, 8 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? · 1 year, 8 months ago

There is no universal way to solve these questions, it just comes down to skill... · 1 year, 8 months ago

Ok. Thanks for clearing! I'll delete my solution now! · 1 year, 8 months ago

An apple a day , keeps the doctor away... It always clears our doubts :P · 1 year, 8 months ago