Question

show that the number m^4+m^3+m^2+m+1 is a square only when m=3 ,m is a natural number

Answer 1

sorry i gotta go and see this later but as a start point note that \[(m^2+\frac{m}{2})^2<m^4+m^3+m^2+m+1<(m^2+\frac{m}{2}+1)^2\]

Answer 2

thanks later

Answer 3

i tried to write it as a square but it cud not work b was always error

Answer 4

(x2+px+1)^2=expression

Answer 5

according to what i said if m is even there is no answer now suppose that m is odd we can have\[m^4+m^3+m^2+m+1=(m^2+\frac{m+1}{2})^2\]which gives m=3

Answer 6

how do we know \[m \neq2k,m=2k+1\]

Answer 7

\[4k^2+k+1\] oh its no real roots

Answer 8

thats waht u have to figure out by ur own :)

Answer 9

\[((2k)^2+\frac{ 2k+1 }{ 2 })^2=(4k^2+k+1/2)^2\]
no solution since
\[1-4(4)(1/2)<0\]

Answer 10

or is it because
the exprexion is the same as\[m(m(m(m+1)+1)+1)+1\]

Answer 11

@mukushla

Answer 12

because it lies between 2 consecutive perfect squares

Answer 13

ofcourse but that is subtle