## 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

1. gorvs Group Title

1+m+m^2(1+m+m^2) $1+m+m ^{2}(m+1)^{2}$

2. gorvs Group Title

sry

3. gorvs Group Title

it will be perfect square for m=0 only

4. John_ES Group Title

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.

5. gorvs Group Title

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

6. shubhamsrg Group Title

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

7. gorvs Group Title

sryyyyyyyyyyyy

8. Loser66 Group Title

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.

9. shubhamsrg Group Title

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

11. mukushla Group Title

really, i cant remember, nice one :)

@mukushla

14. mukushla Group Title

so do u know complete solution now?