$x^3+x+1$=square

[photon]

Find all $x$ non-negative integers, such that $x^3+x+1$ is a perfect square.

[afif mansib ch]

lets assume. \[x^3+x+1=n^2\]
so \[x(x^2+1)=(n+1)(n-1)\]
\[x^2+1\equiv 1(modx)\]
\[\therefore gcd(x^2+1,x)=1\]
let \[x=n+1,x^2+1=n-1\]
so \[n=(x^2+x+1)/2\]
putting the magnitude of n in the 1st eq we get,
\[x^4-2x^3+3x^2-2x-3=0\]
which has no integer solution.

[photon]

let\[x=n+1,x^2+1=n-1\]........

it is not true , $a,b$ are co-prime and $ab=pq$ that doesn't mean a=p or q

[SANZEED]

Here, according to my last calculations,\[x\equiv 0(mod 4),n\equiv 1(mod 2)\] is a case.However I haven't finished yet.

[Fahim Shahriar]

We know n² ≡ 0,1 (mod 4)

For only x ≡ 0 (mod 4), x³+x+1= 0³+0+1= 1 (mod 4)

x³+x+1 = n²
x³+x = n²-1
x(x²+1) = (n+1)(n-1)

For both x≡0 (mod 2) and x≡1 (mod 2), x(x²+1)≡0 (mod 2)
Then (n+1)(n-1) have to be 0 (mod 2)

For n≡0 (mod 2), (n+1)(n-1)≡1 (mod 2)
For n≡1 (mod 2), (n+1)(n-1)≡0 (mod 2) [This one is correct]

So x³+x+1=n² satisfies, when x ≡ 0 (mod 4) And n≡1 (mod 2).