# Help me with this.

Find minimum 2 digit n, such that $(7^{n} + 7^{n-1} + 7^{n-2} + ...... + 7^{3} + 7^{2} + 7^{1} + 7^{0})$ is a perfect square, since known $(7^{3} + 7^{2} + 7^{1} + 7^{0})$ is a perfect square.

Note : $n \neq 3$ !!! Note by Bryan Lee Shi Yang
4 years, 11 months ago

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I am sorry, bcs. I have no answer.

- 4 years, 11 months ago

If you take $n=0$, then you expression will be equal to $1$ which is a perfect square. So $n=0$ can be the one value. Also try to use the latex coding in your notes and problems (now I've put latex in it) , if you need a guide for latex, you can find it just by clicking here .!!! @Bryan Lee Shi Yang

- 4 years, 11 months ago

Up till n=22, only n=0 and n=4 are perfect square according to what I get from TI -83+ calculator. After that I do not get accurate calculations.

- 4 years, 10 months ago

There is no solution < 270.

- 4 years, 10 months ago

Are there any solutions bigger than 270?

- 4 years, 10 months ago

CAN U WRITE A PROOF TO SUPPORT THIS

- 4 years, 10 months ago

$n = 37$ satisfies!

- 4 years, 10 months ago

>> n = 4

>> while (((7(n+1)-1)/6)0.5 - int(((7(n+1)-1)/6)0.5)) != 0:

                     n = n+1


My code! *Easy code, though!

- 4 years, 10 months ago

No it doesn't. 21655801907853836686195357616408 is not a square number.

- 4 years, 10 months ago

@D G $\frac{{7}^{38}-1}{6} = 21655801907853831608163177358336 = {4653579472605344}^{2}$

- 4 years, 10 months ago

http://www.wolframalpha.com/input/?i=%287%5E38+-+1%29%2F6+&dataset=

- 4 years, 10 months ago

There has to be a problem with my python idle, then. Thanks!

- 4 years, 10 months ago

Sorry, it is +1, not -1.

- 4 years, 10 months ago

Oh maybe. My IDLE is just reversing it up. When I replace - with +, it shows the above value(which should come with -).

- 4 years, 10 months ago