1)$1+a+a^2=1+b+b^2+b^3\Rightarrow (a-b)(a+b+1)=b^3$ $p\mid a-b , p\mid a+b+1 \Longrightarrow p\mid 2b+1,p\mid b \Longrightarrow p=1$ $\Longrightarrow a-b=x^3,a+b+1=y^3,xy=b\Longrightarrow b=\frac{y^3-x^3-1}{2}=xy$ $\Longrightarrow (y-x)(y^2+xy+x^2)=2xy+1$ but $(y-x)(y^2+xy+x^2)>2xy+1$

How did you get that $p|b$?

mousavi wrote: ...$p\mid a-b , p\mid a+b+1 \Longrightarrow p\mid 2b+1,p\mid b \Longrightarrow p=1$... Better: ...Let $d=\gcd(a-b, a+b+1)$. If $d>1$, then $d$ has a prime divisor, say $p$. So $p \mid b^3$, which means that $p \mid b$. Now since we got $p \mid 2b+1$, we have $p \mid 1$, a contradiction. So $d=1$...

Obel1x wrote: Let $n$ be a natural number, for which we define $S(n)=\{1+g+g^2+...+g^{n-1}|g\in{\mathbb{N}},g\geq2\}$ $a)$ Prove that: $S(3)\cap S(4)=\varnothing$ $b)$ Determine: $S(3)\cap S(5)$ b: let $gcd(a,b)=d>1$ in $a^2+a=b^4+b^3+b^2+b$ and in $(a-b)(a+b+1)=b^3(b+1)$ so let $a=dm,b=dn$ and we know $gcd(m,n)=1$ and $(a-b)(a+b+1)=b^3(b+1)$ becomes $(m-n)(dm+dn+1)=d^2n^3(dn+1)$ where we see $d^2|m-n$ $m\equiv n \pmod{d^2}$ and $ggd(d,m)=ggd(d,n)=1$ let $m=d^2k+n$ we get $a^2+a=b^4+b^3+b^2+b$ becomes somewhere we heve to prove $d=1$ and then we see fast that $2,5$ is only couple. (had submitted with litle mistake)

@SCP: sorry, I don't understand how you finished your proof.

b) $a^2+a=b^4+b^3+b^2+b$ $a=b^2+m$ then $2b^2m+m^2+m=b^3+b$ solving for $m$, the discriminant has to be a square then: $(2b^2+1)^2+4(b^3+b)=k^2$ $4b^4+4b^3+4b^2+4b+1=k^2$ replacing $k=2b^2+b+n$ for $n \in \mathbb{N}$ because $(2b^2+b)^2=4b^4+4b^3+b^2$ $4b^2+4b+1=4b^2n+b^2+2bn+n^2$ $3b^2+4b+1=4b^2n+2bn+n^2$ $b^2(4n-3)+b(2n-4)+n^2-1=0$ Notice that if $n\ge 2$ then $b^2(4n-3)+b(2n-4)+n^2-1>0$ contradiction then $n=1$ $b^2-2b=0$ then $b=2$ and replacing getting $a=5$ Is it ok?

mszew wrote: b) $a^2+a=b^4+b^3+b^2+b$ $a=b^2+m$ then $2b^2m+m^2+m=b^3+b$ solving for $m$, the discriminant has to be a square then: $(2b^2+1)^2+4(b^3+b)=k^2$ $4b^4+4b^3+4b^2+4b+1=k^2$ replacing $k=2b^2+b+n$ for $n \in \mathbb{N}$ because $(2b^2+b)^2=4b^4+4b^3+b$ $4b^2+4b+1=4b^2n+b^2+2bn+n^2$ $3b^2+4b+1=4b^2n+2bn+n^2$ $b^2(4n-3)+b(2n-4)+n^2-1=0$ Notice that if $n\ge 2$ then $b^2(4n-3)+b(2n-4)+n^2-1>0$ contradiction then $n=1$ $b^2-2b=0$ then $b=2$ and replacing getting $a=5$ Is it ok? Yes it is OK. @ Amparvardi: I'd made some mistake, hence it wasn't complete.

mousavi wrote: 1)$1+a+a^2=1+b+b^2+b^3\Rightarrow (a-b)(a+b+1)=b^3$ $p\mid a-b , p\mid a+b+1 \Longrightarrow p\mid 2b+1,p\mid b \Longrightarrow p=1$ $\Longrightarrow a-b=x^3,a+b+1=y^3,xy=b\Longrightarrow b=\frac{y^3-x^3-1}{2}=xy$ $\Longrightarrow (y-x)(y^2+xy+x^2)=2xy+1$ but $(y-x)(y^2+xy+x^2)>2xy+1$ Why does $(y-x)(y^2+xy+x^2)>2xy+1$? Or, can we encounter another contradiction which is easier to see?