Question

for how many integers \(m\) the expression\[1+m+m^2+m^3+m^4\]becomes a perfect square?

Answer 1

1+m+m^2(1+m+m^2)
\[1+m+m ^{2}(m+1)^{2}\]

Answer 2

sry

Answer 3

it will be perfect square for m=0 only

Answer 4

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.

Answer 5

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

Answer 6

you have done a mistake in your 2nd step @gorvs

Answer 7

sryyyyyyyyyyyy

Answer 8

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.

Answer 9

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

Answer 10

i asked you this before @mukushla

Answer 11

really, i cant remember, nice one :)

Answer 12

http://openstudy.com/users/jonask#/updates/50783e67e4b02f109be41dff

Answer 13

@mukushla

Answer 14

so do u know complete solution now?

Answer 15

well somehow i gave up and jus considered the fact that it is between two perfect squares so m=3