Proving Question (1 Viewer)

hunterbear · Sep 7, 2013

Prove that there are no integers a, b and c that satisfy the equation

a³+b³=7c³+3

SpiralFlex · Sep 7, 2013

$Suppose$ \;a, b\; $and$ \;c \;$are integers. We have,$

$a^3+b^3$

$We examine$\;a^3\;$and we notice that,$

$a^3 \equiv 0\; ($modulo$\; 7)$
$a^3 \equiv 1\; ($modulo$\; 7)$
$a^3 \equiv 6 \;($modulo$\; 7)$

$This means, by addition in modular arithmetic,$

$a^3+b^3 \equiv 0 \;($modulo$\; 7)$
$a^3+b^3 \equiv 1 \;($modulo$\; 7)$
$a^3+b^3 \equiv 2 \;($modulo$\; 7)$
$a^3+b^3 \equiv 5\; ($modulo$\; 7)$
$a^3+b^3 \equiv 6 \;($modulo$\; 7)$

$Also notice,$

$7c^3 $\;is an integer, so$\; 7c^3 + 3\; $is a multiple of$\; 7\; $with$\; 3\; $added on. Hence,$$

$7c^3+3 \equiv 3 \;($modulo$\; 7)$

$Thus, the equation is never satisfied.$

hunterbear · Sep 7, 2013

Is there any simpler way of proving this?

HSC2014 · Sep 7, 2013

What is modulus 7 lol. Seems interesting

SpiralFlex · Sep 7, 2013

HSC2014 said:
What is modulus 7 lol. Seems interesting

This article may interest you then:

http://www.artofproblemsolving.com/Wiki/index.php/Modular_arithmetic/Introduction

hunterbear · Sep 7, 2013

So... Is there any other way? I just find that modular arithmetic generally confuses me :s

RealiseNothing · Sep 7, 2013

Now try the question for:

$a^n + b^n = c^n$

$n>2$ and is an integer.

SpiralFlex · Sep 7, 2013

hunterbear said:
So... Is there any other way? I just find that modular arithmetic generally confuses me :s

Where did you get this question? I can only see anyway I select to argue this question is always through modular arithmetic (and perhaps divisibility)

hunterbear · Sep 7, 2013

I found this in some old book for some competition at where I go tutoring.

Andrew Wiles · Sep 7, 2013

RealiseNothing said:
Now try the question for:

$a^n + b^n = c^n$

$n>2$ and is an integer.

Iill get back to you shortly with the proof.

RealiseNothing · Sep 7, 2013

Andrew Wiles said:
Iill get back to you shortly with the proof.

this just happened hahahahah

HSC2014 · Sep 7, 2013

Hahaha. 7 years later...

asianese · Sep 7, 2013

omg LOL

braintic · Sep 7, 2013

I'm just wondering:

Are you all laughing because you think the guy is serious?

Or are you laughing because you realised his name (Andrew Wiles) is the name of the person who finally proved Fermat's Last Theorem about 20 years ago.

RealiseNothing · Sep 7, 2013

braintic said:
I'm just wondering:

Are you all laughing because you think the guy is serious?

Or are you laughing because you realised his name (Andrew Wiles) is the name of the person who finally proved Fermat's Last Theorem about 20 years ago.

the latter, hence why the guy above said "7 years later"

HSC2014 · Sep 7, 2013

Mhm

Andrew Wiles went into solitude for 6 years to work on the conjecture before releasing a proof that had an error in it, requiring an additional year to correct. (or something like that)

Kobweb · Sep 8, 2013

Funniest thread NA

obliviousninja · Sep 8, 2013

Bahahaha.

On another note, is this a HSC-worthy question?

SpiralFlex · Sep 8, 2013

No. Which was why it was posted in the extracurricular topics.

seanieg89 · Sep 8, 2013

It's a shame. I think elementary number theory would be an excellent thing to teach high school students.

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