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

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TopNewestI am sorry, bcs. I have no answer.

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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

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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.

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There is no solution < 270.

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Are there any solutions bigger than 270?

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CAN U WRITE A PROOF TO SUPPORT THIS

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\(n = 37\) satisfies!

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My code! *Easy code, though!

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

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