There are no positive integral solutions, because (in $\mathbb{Z}[\sqrt{-2}]$, which has FTA) we need to solve $x^3=(1+y\sqrt{-2})(1-y\sqrt{-2})$. The LHS factors are coprime so $(a+b\sqrt{-2})^3=1+y\sqrt{-2}$. Equating terms gives $a^3-6ab^2=1\Rightarrow(a,b)=(1,0)$ so $x^3=\left((1+0\sqrt{-2})(1-0\sqrt{-2})\right)^3=1$ and $y=0$. But $0$ is not positive.

wrong sol.

yayitsme wrote: $(x-1)(x^2+x+1)=2y^2$ Case 1: $(x-1,x^2+x+1)=1$ Then $x^2+x+1=a^2$ and $x-1=2b^2$ which implies $4b^4+6b^2-3=a^2$ Easy to see thats not possible. Case 2:Their gcd is $3$. In that case $x^2+x+1=3a^2$ and $x-1=6b^2$ which $\implies a^2=12b^4+9b^2+3$ which is impossible since $3 \mid \not 4b^4+3b^2+1$(does not divide) Hence no such positive integer exists. In the case 2 i think that we have $a^2=12b^4+6b^2+1$ and not $a^2=12b^4+9b^2+3$

grupyorum wrote: Solve the following equation in positive integers: $x^{3} = 2y^{2} + 1 $ As Vworld: We are going to work $\mathbb{Z}[\sqrt{-2}]$: First it is obvious that $x=odd$. Factory the equation as: $x^3=(1+y\sqrt{-2})(1-y\sqrt{-2})$. Claim:$(1+y\sqrt{-2})$ and $(1-y\sqrt{-2}$ are comprime in $\mathbb{Z}[\sqrt{-2}]$: Let $d=((1+y\sqrt{-2},(1-y\sqrt{-2})$ then $d$ device addition so $d|2$ but also $d|x$ and we know that $x=odd$ so $d=1$ This means that there is $a,b$ integers such that: so $(a+b\sqrt{-2})^3=1+y\sqrt{-2}$.this give as the following system: $a^3-6ab^2=1$ $3a^2b-4b^3=y$ In the first we have $a|1$ so $a=1or-1$ . For $a=-1$ we have $6b^2=2$ no solution For $a=1$ we have $6b^2=0$ so $b=0$ which give that $y=0$ and $x=1$ So no solution in positive integer. Let me know if i am wrong

can anyone tell me what is $\mathbb{Z}[\sqrt{-2}]$ ?

Ucchash wrote: can anyone tell me what is $\mathbb{Z}[\sqrt{-2}]$ ? Check here.

Prod55 Wrong solution (loxmisanyo shuni sezmagan boseng)

Clearly $x$ is odd and we have $x^3 = (1 + y\sqrt{-2})(1-y\sqrt{-2})$. If $d$ is a common divisor of the two factors on the left, then $d$ divides $2$, so its norm must divide $8$. But this norm cannot be even since $d$ also divides $y^3$, which has the odd norm $y^6$. Therefore $x\pm\sqrt{-2}$ are coprime and since all units $\pm 1 = (\pm 1)^3$ are cubes in $\mathbb{Z}[\sqrt{-2}]$, it must be the case that $1+y\sqrt{-2}=(a+b\sqrt{-2})^3$ for some integers $a$ and $b$. Opening the brackets gives $1 = a^3 - 6ab^2 = a(a^2-6b^2)$ and $y = 3a^2b - 2b^3$, hence $a=\pm 1$, $b=0$ and $y=0$. Therefore $(1,0)$ is the only solution.

grupyorum wrote: Solve the following equation in positive integers: $x^{3} = 2y^{2} + 1 $ In fact, we can solve the generalized equation $x^n=2y^2+1$ where $n \geqslant 3$.

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[url]"C:\Users\hp\Pictures\Capture d’écran 2023-05-20 180314.png"[/url]

No solution. We have $(x^2-x+1)(x+1)=x^3+1=2(y^2+1)$ We know that $p|LHS \implies p \equiv 0,1(mod 4)$ so $x^2-x+1 \equiv 1 (mod 4) \implies x\equiv 1 (mod 4)$ Let's say $x=4k+1$ \[64k^3+48k^2+12k+1=x^3=2y^2+1\]\[64k^3+48k^2+12k=2y^2\]\[2k(16k^2+12k+3)=y^2\]$v_2(LHS)=v_2(2k)=1+v_2(k)=v_2(RHS)=2v_2(y)$ Let $k=2^{2q-1}m$ such that $m$ is odd.$y=2^qn$ and $q \geq 1$ \[2^{2q}(2^{4q}m^2+3.2^{2q+1}m+3)=2^{2q}n^2\]\[2^{4q}m^2+3.2^{2q+1}m+3=n^2\]\[(2^{2q}m+3)^2=2^{4q}m^2+3.2^{2q+1}m+9=n^2-6\]\[\implies 6=n^2-(2^{2q}m+3)^2\]Both $n$ and $2^{2q}m+3$ should be odd so $4|RHS=6$ Contradiction

Guys this problem easy Just look to equation $8x^3=16y^2+8$ $8x^3-8=16y^2=k^2$ $(|\sqrt{8x^3}|)^2>8x^3-8>(|\sqrt{8x^3}|-1)^2$