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mukushla
Group Title
for how many integers \(m\) the expression\[1+m+m^2+m^3+m^4\]becomes a perfect square?
 one year ago
 one year ago
mukushla Group Title
for how many integers \(m\) the expression\[1+m+m^2+m^3+m^4\]becomes a perfect square?
 one year ago
 one year ago

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gorvs Group TitleBest ResponseYou've already chosen the best response.0
1+m+m^2(1+m+m^2) \[1+m+m ^{2}(m+1)^{2}\]
 one year ago

gorvs Group TitleBest ResponseYou've already chosen the best response.0
it will be perfect square for m=0 only
 one year ago

John_ES Group TitleBest ResponseYou've already chosen the best response.0
Well, numerically I find others m not zero, m=1,m=0 and m=3. A more rigorous proof is needed for all m in integers.
 one year ago

gorvs Group TitleBest ResponseYou've already chosen the best response.0
1+m(m+1)+m^3(m+1)=1+(m+1)(m+m^3) =1+(m+1)(m^2(m+1)) =1+(m^2)*(m+1)^2 =1+(m(m+1))^2 except zero it will not be a perfect square bcoz after sqauring 1is added which makes it a perfect square
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.1
you have done a mistake in your 2nd step @gorvs
 one year ago

gorvs Group TitleBest ResponseYou've already chosen the best response.0
sryyyyyyyyyyyy
 one year ago

Loser66 Group TitleBest ResponseYou've already chosen the best response.0
all you are off line, I can write whatever I like, I choose (m^2 +1) ^2 as my perfect square, so \[(m^2+1)^2=m^4+2m^2+1\] to find out the solution, I let \[m^4+2m^2+1=m^4+m^3+m^2+m+1\] so, when \[m^2=m+m^3\], I have perfect square \[m^2=m^2(\frac{1}{m}+m)\] > \[\frac{1}{m}+m=1\] \[m^2m+1=0\]no real solution for this.
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.1
I got a method but not too sure if its ideal lets take 1+ m + m^2 + m^3 + m^4 = p^2 for some integer p then m(1+m) + m^3 (1+m) = (p+1) (p1) m(1+m)(1+m^2) = (p+1)(p1).1 now comparing will get our answers, but it'll be bit lengthy as too many permutations possible like this one yields m=3 and p=11 m(1+m) = 1+p (1+m^2) = p1 like wise we will make all possible comparisons and try to find out integer solutions
 one year ago

Jonask Group TitleBest ResponseYou've already chosen the best response.0
i asked you this before @mukushla
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.0
really, i cant remember, nice one :)
 one year ago

Jonask Group TitleBest ResponseYou've already chosen the best response.0
http://openstudy.com/users/jonask#/updates/50783e67e4b02f109be41dff
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.0
so do u know complete solution now?
 one year ago

Jonask Group TitleBest ResponseYou've already chosen the best response.0
well somehow i gave up and jus considered the fact that it is between two perfect squares so m=3
 one year ago
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