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.

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).

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.

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.

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 \(2\)".

Easy Math Editor

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boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

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## Comments

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

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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 \( 1^2 + 8^2 = 4^2 + 7^2\).

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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.

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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).

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

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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.

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Nice question

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

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

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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.

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

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Oooops, yes I do. Edited. Thanks!

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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 \(\boxed{1}\).

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

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Comment deleted Nov 20, 2015

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

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

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@Aditya kumar ,Interesting it really works.Is this method always applicable in such problems.

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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 \(2\)".

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