If n is any positive integer show that the integral part of $$(3+\sqrt7)^n$$is a odd number

I have no idea how to begin this problem but it is given in the chapter of binomial theorem so I hope that it is found using that only

  • $\begingroup$ Have you tried using the binomial theorem? How far do you get? Why are you stuck? $\endgroup$ – saulspatz Jan 20 at 16:57
  • $\begingroup$ I have tried but haven't got to a useful reasult $\endgroup$ – user735102 Jan 20 at 16:59
  • 3
    $\begingroup$ Asking for upvotes is more reason for me not to upvote. If you want attention give an actual reason for attention. Why does anyone care about this problem? What information from the binomial theorem do you have? etc. $\endgroup$ – Simply Beautiful Art Jan 20 at 17:01
  • $\begingroup$ Sorry sir I would not repeat this in anytime in near future $\endgroup$ – user735102 Jan 20 at 17:03


Consider $(3+\sqrt7)^n+(3-\sqrt7)^n$

  • $\begingroup$ +1, but why is it true the identity with the question of the user? $\endgroup$ – Sebastiano Jan 20 at 18:10
  • 1
    $\begingroup$ @Sebastiano: binomial expansion shows the sum is an even integer. $(3-\sqrt7)^n<1$, so $\lfloor(3+\sqrt7)^n\rfloor$ is odd $\endgroup$ – J. W. Tanner Jan 20 at 18:20

'I'to denote the integral and 'f'to denote the fractional part of $(3+√7)^n$

Now $(3-√7)^n$ is less than 1 and a proper fraction let's denote it by f'

$(3+√7)^n=3^n+ C_13^{n-1}√7......$


As you can see when we add them the irrational terms cancel out.

$(3+√7)^n+(3-√7)^n$=I+f+f'= even integer

But since f and f' are proper fractions there are some must be 1

Hence we conclude that it's integral part is odd.

  • $\begingroup$ This is what I did $\endgroup$ – user735102 Jan 20 at 17:16
  • $\begingroup$ Any corrections needed? $\endgroup$ – user735102 Jan 20 at 17:17
  • 1
    $\begingroup$ It would be instructive to explain in detail why $(3+\sqrt7)^n+(3-\sqrt7)^n$ is an even integer $\endgroup$ – J. W. Tanner Jan 20 at 17:20
  • $\begingroup$ Thank you sir I will edit my answer $\endgroup$ – user735102 Jan 20 at 17:22
  • 1
    $\begingroup$ Did you mean their sum when you wrote there are some? Also, I still think you should explain better why it's an even integer $\endgroup$ – J. W. Tanner Jan 20 at 18:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy