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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, 2 months ago

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I am sorry, bcs. I have no answer. · 2 years, 2 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 · 2 years, 2 months ago

CAN U WRITE A PROOF TO SUPPORT THIS · 2 years, 2 months ago

There is no solution < 270. · 2 years, 2 months ago

@D G Are there any solutions bigger than 270? · 2 years, 2 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. · 2 years, 2 months ago

$$n = 37$$ satisfies! · 2 years, 2 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! · 2 years, 2 months ago

No it doesn't. 21655801907853836686195357616408 is not a square number. · 2 years, 2 months ago

@D G $$\frac{{7}^{38}-1}{6} = 21655801907853831608163177358336 = {4653579472605344}^{2}$$ · 2 years, 2 months ago

Sorry, it is +1, not -1. · 2 years, 2 months ago

Oh maybe. My IDLE is just reversing it up. When I replace - with +, it shows the above value(which should come with -). · 2 years, 2 months ago

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