Prove that either $(2^{10500} + 15)$ or $(2^{10500} + 16)$ is not a perfect square.
how should I solve this problem? what is the idea for solving this kind of problems?
Thank you so much
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1Hint: you don't need to compute anything except $2^{10500} > 0$. – Henry Swanson Dec 06 '13 at 05:41
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when $a^2-b^2=1?$ – lab bhattacharjee Dec 06 '13 at 05:45
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1Hint: both of them are between $n^2$ and $(n+1)^2$ for some $n$. – Du Phan Dec 06 '13 at 05:45
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Shouldn't this say "neither/nor" rather than "either/or" ? – DanielV Dec 06 '13 at 06:38
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It is clearly a problem that is designed to intimidate you, so that you miss the absolutely simple solution. You solve this by ignoring the homungous numbers that are intended to frighten you and turn your brain into a mush, and realise that the only two consecutive perfect squares are 0^2 and 1^2. – gnasher729 May 18 '14 at 11:19
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Hint: What is the next perfect square above $2^{10500}$? How does that compare to the numbers you're given?
John
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Find square root of $2^{10500}$: $$(2^{10500})^\frac12=2^{10500\times\frac12}=2^{5250}$$ and the next square is $$(2^{5250}+1)^2=2^{10500}+2^{5251}+1$$ The difference between these $2$ squares is $$2^{10500}+2^{5251}+1-2^{10500}=2^{5251}+1$$ $2^{5251}+1$ is way bigger than $15$ or $16$
Edit: This proves none of them are perfect squares. The difference between the $2$ numbers you have given us is only $1$ and no 2 perfect squares except 0 and 1 have a difference of 1.
Abraham Zhang
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The difference between two squares $n^2$ and $(n + a)^2$ is $2an + a^2$.
Say they are both squares, $n^2 = 2^{10500} + 15$ and $(n+a)^2 = 2^{10500} + 16$. Taking their difference, $1 = 2an + a^2$. But $n$ is huge and $a^2$ is positive, so that's absurd.
Henry Swanson
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