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 : $n \neq 3$ !!!

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Comments

I am sorry, bcs. I have no answer.

Bryan Lee Shi Yang - 6 years, 3 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

Sandeep Bhardwaj - 6 years, 3 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.

Niranjan Khanderia - 6 years, 3 months ago

There is no solution < 270.

D G - 6 years, 3 months ago

Are there any solutions bigger than 270?

Bryan Lee Shi Yang - 6 years, 3 months ago

CAN U WRITE A PROOF TO SUPPORT THIS

vishwesh agrawal - 6 years, 3 months ago

$n = 37$ satisfies!