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 $a^2-b^2+c^2$

#Algebra #PreRMO

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

Markdown	Appears as
`italics` or `_italics_`	italics
`bold` or `__bold__`	bold
- bulleted - list	bulleted list
1. numbered 2. list	numbered list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1 paragraph 2	paragraph 1 paragraph 2
`[example link](https://brilliant.org)`	example link
`> This is a quote`	This is a quote
# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"	# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"

Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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 - 5 years, 6 months ago

Nice question

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

Calvin Lin Staff - 5 years, 6 months ago

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

Siddhartha Srivastava - 5 years, 6 months ago

Oooops, yes I do. Edited. Thanks!

Calvin Lin Staff - 5 years, 6 months ago

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

Aditya Kumar - 5 years, 6 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.

Calvin Lin Staff - 5 years, 6 months ago

@Calvin Lin – Wow! Nice way!

Aditya Kumar - 5 years, 6 months ago

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

Tejas Khairnar - 5 years, 6 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.

Calvin Lin Staff - 5 years, 6 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.

Nihar Mahajan - 5 years, 6 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).

Calvin Lin Staff - 5 years, 6 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 $1^2 + 8^2 = 4^2 + 7^2$ .

Calvin Lin Staff - 5 years, 6 months ago

2^{0}

Saroja Gudimetla - 2 years, 10 months ago

Kuhu Raychaudhuri - 2 years, 10 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 $\boxed{1}$ .

Siddharth Singh - 5 years, 6 months ago

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

Julian Poon - 5 years, 6 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$