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# proof for equation with no solution

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1. How to prove that there is no solution for this equation in natural numbers:
2. Consider modulo arithmetic.

I could tell you what base works but not the justification for choosing it, as my number theory is too rusty.

Is this really "Secondary school" level?
3. Could you not use the quadratic formula?

I then wouldn't really know where to go from there, sorry.
I just remember seeing something like this a couple of ears ago.
4. But quadratic formula is not proving anything i think ,i know that we can prove it by modulo but how ?
5. (Original post by MAA_96)
i know that we can prove it by modulo but how ?
With the right base, the LHS can only be equal to certain values, and the RHS can only be equal to certain other values; and those two sets of values do not intersect, and hence there is no solution.

Base is

Spoiler:
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9
6. 3n^2+3n+7 =1 mod 3. Thus K^3 must be 1 mod 3. let K=3m+1 imples
3n^2+3n+7=27(m^3+m^2)+9m+1
n^2+n+2=9(m^3+m^2)+3m
rhs=0 mod 3. n^2+n+2 is either 2 or 1 mod 3, never 3 by considering residue classes
done
7. (Original post by Blutooth)
3n^2+3n+7 =1 mod 3. Thus K^3 must be 1 mod 3. let K=3m+1 imples
3n^2+3n+7=27(m^3+m^2)+9m+1
n^2+n+2=9(m^3+m^2)+3m
rhs=0 mod 3. n^2+n+2 is either 2 or 1 mod 3, never 3 by considering residue classes
done
Clever. I wasn't thinking.
8. (Original post by wcp100)
Clever. I wasn't thinking.
Thank you, but if you want clever I just solved an IMO problem in the summer maths thread. Took me 2 days to work out and I was thinking
9. (Original post by Blutooth)
...
Nice; for some reason I was getting 2 when thinking of 7 (mod 3) ; brain dead.

Defo. not "secondary school".

+rep.

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