Fermat's last theorem

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Fermat's last theorem

Fermat's last theorem claims that A^n + B^n = c^n has no solution if n is bigger than 2.
A student claims that he has proved 681^11 + 309^11 = 876^11.
Could you convince the student that he is wrong? Using a hand calculator only if need be.

Posted 213 day ago

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matherapist	Best Answer Yes, I could convince the student he is wrong. any integer # ending in 1 raised to any power ends in 1 any integer # ending in 9 raised to a power ends in 1 or 9 summing these will end in 2 or 0 and integer # ending in 6 raised to any power ends in 6 Thus the equality will not hold. So Raj, do you pose these problems for a reason? Posted 213 day ago ( 1 ) ( 0 ) Report Abuse

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Answers (10)

atta

Actually, it has proven by 17th century mathematicians that there is a solution to numbers with at least two. But that which has never been proven is numbers greater than 5. Two 17th century mathematicians invented advanced algebraic methods for finding solutions to polynomial degree powers of 3 but no so with 5. Thus, Fermat's Last Theorem claims that x^n + y^n = z^n has no solution for any real-number n >= 5.

J.C

Posted 155 day ago

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jayteacher

Sure, math therapist, you run into rounding errors, however, the 2 sides of the equal sign have such a large difference in values that the rounding errors are insignificant.

Posted 209 day ago

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matherapist

even with scientific notation, you will run into rounding problems due to a limit on significant figures for this type of problem. Perhaps a double precision 64 bit machine would work, but my calculator doesn't have that capability.

Posted 211 day ago

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jayteacher

I scrolled to the bottom of the page, and just read that you were a math professor at Cal State. My apologies to your comment. Of course you know that a scientific notation calculator is needed to discover the inequality that you gave in Fermat's Last Theorem.

Posted 211 day ago

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jayteacher

Try a calculator with scientific notation capability.

Posted 211 day ago

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matherapist

Raj

I retired from NASA after 40 years. I thot that this board was to answer ??? from students. I doubt if students try to answer any of these. It would be nice if they did. Meanwhile my Bee problem goes unanswered. I think we'll find other tutors will answer the questions just to show they can, or can't.

M

Posted 213 day ago

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This note is for jayteacher to read:
A hand held calculator will not be able to do what you suggest! rajbaba78

Posted 213 day ago

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jayteacher

Have you tried entering the values of exponential terms on the left in a calculator and adding; then finding out if the answer is the same value as the exponential term on the right?

I think the student should try this before asking for help.

Posted 213 day ago

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matherapist asked me why I post this type of questions?
The idea is for students to use calculator and see how numbers behave when raised to different powers.
They will find out that any number ending in 1, raised to any power will have a last digit of 1, and similarly, numbers ending in 9, will have a last digit of either 1 or 9, also numbers ending in 6, will have a last digit of 6. Therefore they will be able to answer the question easily.
I have retired as a mathematics professor from California State University, and I am using my experience to teach students how to think and how to solve mathematical problems. Good luck matherapist!
You did a good job !!

Posted 213 day ago

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matherapist

Yes, I could convince the student he is wrong.

any integer # ending in 1 raised to any power ends in 1
any integer # ending in 9 raised to a power ends in 1 or 9

summing these will end in 2 or 0

and integer # ending in 6 raised to any power ends in 6

Thus the equality will not hold.

So Raj, do you pose these problems for a reason?

Posted 213 day ago

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