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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
2 years ago

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I am sorry, bcs. I have no answer. Bryan Lee Shi Yang · 2 years ago

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@Bryan Lee Shi Yang 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 · 2 years ago

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CAN U WRITE A PROOF TO SUPPORT THIS Vishwesh Agrawal · 2 years ago

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There is no solution < 270. D G · 2 years ago

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@D G Are there any solutions bigger than 270? Bryan Lee Shi Yang · 2 years ago

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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. Niranjan Khanderia · 2 years ago

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\(n = 37\) satisfies! Kartik Sharma · 2 years ago

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@Kartik Sharma

>> 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! Kartik Sharma · 2 years ago

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@Kartik Sharma No it doesn't. 21655801907853836686195357616408 is not a square number. D G · 2 years ago

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@D G \(\frac{{7}^{38}-1}{6} = 21655801907853831608163177358336 = {4653579472605344}^{2}\) Kartik Sharma · 2 years ago

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@Kartik Sharma Sorry, it is +1, not -1. Bryan Lee Shi Yang · 2 years ago

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@Bryan Lee Shi Yang Oh maybe. My IDLE is just reversing it up. When I replace - with +, it shows the above value(which should come with -). Kartik Sharma · 2 years ago

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@Kartik Sharma http://www.wolframalpha.com/input/?i=%287%5E38+-+1%29%2F6+&dataset= Siddhartha Srivastava · 2 years ago

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@Siddhartha Srivastava There has to be a problem with my python idle, then. Thanks! Kartik Sharma · 2 years ago

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