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mukushla
for how many integers \(m\) the expression\[1+m+m^2+m^3+m^4\]becomes a perfect square?
1+m+m^2(1+m+m^2) \[1+m+m ^{2}(m+1)^{2}\]
it will be perfect square for m=0 only
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.
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
you have done a mistake in your 2nd step @gorvs
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 \[m^2=m^2(\frac{1}{m}+m)\] ---> \[\frac{1}{m}+m=1\] \[m^2-m+1=0\]no real solution for this.
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) (p-1) m(1+m)(1+m^2) = (p+1)(p-1).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) = p-1 like wise we will make all possible comparisons and try to find out integer solutions
i asked you this before @mukushla
really, i cant remember, nice one :)
http://openstudy.com/users/jonask#/updates/50783e67e4b02f109be41dff
so do u know complete solution now?
well somehow i gave up and jus considered the fact that it is between two perfect squares so m=3