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Prove $\sqrt{2013^{2016}+2014^{2016}}$ is irrational.

I've looked at the proofs for proving $\sqrt{\mathstrut 2}$, $\sqrt{\mathstrut 3}$, and$\sqrt{\mathstrut 15}$ irrational. Using proof by contradiction, assuming this number is rational and in the form of $\frac{m}{n}$ where m&n are relatively prime. I understand that for the $\sqrt{\mathstrut 2}$ proof you essentially get that they're both even and thus not relatively prime, and $\sqrt{\mathstrut 3}$ proof involves rewriting m and n as 2a+1 and 2b+1 and getting LHS odd and RHS even. For the $\sqrt{\mathstrut 15}$ you get the factors of 3 and 5 being odd on one side and even on the other.

I'm having difficulty using any of these techniques to prove the problem is irrational. Is there a way I'm supposed to rewrite 2013 and 2014 in order to simplify this problem into one of the above mention proofs?

Any thoughts or suggestions would be extremely helpful. Thank you!

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What are the possible last digits in base 10 for square numbers?

What is the last digit of the radicand? (bit inside the root)

Let me know if you need further help.

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  • $\begingroup$ What about rest from division by $5$ instead? May be less troublesome $\endgroup$ – Jakobian Jan 17 at 12:02
  • $\begingroup$ Can this number be divided by 5? $\endgroup$ – Ben Crossley Jan 17 at 12:04
  • $\begingroup$ Instead of division by $10$, let's do division by $5$. Then finding possible rests of squares will be easier $\endgroup$ – Jakobian Jan 17 at 12:04
  • $\begingroup$ Ahh, I see. Yes that would work, seems sensible to stay in the base it's written in to me :) $\endgroup$ – Ben Crossley Jan 17 at 12:05
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    $\begingroup$ Edit - Thank you for the start - so only 1, 4, 9, 6 and 5. And since you would get last digits of 1 and 6 for 2013 and 2014, your last digit would be 7 in the radicand, and since 7 is not one of the 5 numbers above, it cannot be a perfect square. Thanks! $\endgroup$ – Abe Cain Jan 17 at 12:13
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$$ 2013^{2016}+2014^{2016}\equiv 2\pmod{5} $$ (by Fermat's little theorem) implies that $ 2013^{2016}+2014^{2016}$ is not a square, hence its square root is irrational.

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Or, for a different approach, a sum of two perfect fourth powers is never a perfect square unless they're $0$ and $1$. Reference - this is a theorem that Fermat actually proved.

And, of course, the square root of any positive integer that isn't a perfect square is irrational.

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