# Is there a way to find the integer solutions of $x^2 = y^3 + 1$ with elementary NT [duplicate]

I am a 14 year old that started preparing for the youth Balkaniad. I stumbled on some questions that will be way easier if only I knew how to solve this( I am not allowed to use Catalans in the competition)

So far I have expressed it to $$(x+1)(x-1) = y^3$$ and further proving that gcd(x+1,x-1) = 1,2 so we have either that $$x+1 = a^3$$, $$x-1 = b^3$$ ( I already solved it when x+1, x-1 are coprime) or when $$x+1 = 2a^3$$, $$x-1 = 2b^3$$(I am struggling with this part)

The solutions that I have found so far are 2 and 3 aka Catlans $$3^2 - 2^3 = 1$$

Any help will be well appreciated

• Please visit :math.stackexchange.com/questions/1898338/… Commented Aug 24, 2022 at 20:55
• You want $x+1=2a^3, x-1=4b ^3$ or $x+1=4a^3, x-1=2b^3.$ If $x+1=2a^3$ and $x-1=2b^3,$ then $(x-1)(x+1)$ is not a cube. Commented Aug 24, 2022 at 20:58
• So you get the equations $a^3-2b^3=1$ and $2a^3-b^3=1.$ Commented Aug 24, 2022 at 21:00
• This is from Dickson's History of the Theory of Numbers, vol. 2, p 533. L. Euler$^4$ proved that x3+ 1 = 0 has no [positive] rational solution except x=2. The reference is Comm. Acad. Petrop., 10, 1738, 145; Comm. Arith. Coll., I, 33-34; Opera Omnia, (1), II, 56-58. Proof republished by E. Waring, Medit. Algebr., ed. 3, 1782, 374-5. Commented Aug 25, 2022 at 4:00
• The solution you have not pursued for $(x-1)(x+1)=y^3$ is $x-1=y,\ x+1=y^2$ which by subtraction leads to $y^2-y=y(y-1)=2$. Since $2$ is prime, you quickly see $y-1=1$ and $y=2$ is the required solution. Commented Aug 25, 2022 at 16:20