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I will present an elementary solution for this fundamental Mordell equation.We are looking for integral solutions of the equation:[math]a^2+1 = b^3 => b^3-1 = a^2 … (1)[/math]Since the LHS of [math](1)[/math] is nonnegative integer, so must be the RHS of [math](1)[/math]. Hence, we are only looking for nonnegative integral solutions of [math](1)[/math].Now, [math](1)[/math] takes the form:[math](b-1)(b^2+b+1) = a^2[/math]Let us assume that [math]b[/math] is odd. Then, [math]a[/math] is even and there exist positive integers [math]k[/math], [math]m[/math], such that:[math]b = 2k+1 … (2)[/math][math]a = 2m … (3)[/math]By [math](1)[/math], [math](2)[/math] and [math](3)[/math], we take:[math](2k+1)^3-1 = (2m)^2 => 8k^3+12k^2+6k+1-1 = 4m^2 => [/math][math]8k^3+12k^2+6k = 4m^2 => 4(2k^3+3k^2)+6k = 4m^2 … (4)[/math]Now, since the RHS of [math](4)[/math] is [math]0mod4[/math], so must be the LHS of it. Hence, [math]k[/math] is even, which means that there exists positive integer [math]p[/math], such that [math]k = 2p[/math]. Hence, [math](4)[/math] takes the form:[math]4(16p^3+12p^2+3p) = 4m^2 => 16p^3+12p^2+3p = m^2 =>[/math][math]p(16p^2+12p+3) = m^2 … (5)[/math]Evidently [math]gcd(p, 16p^2+12p+3) = 1[/math] or [math]3[/math]In the case at which [math](p, 16p^2+12p+3) = 1[/math], by [math](5)[/math] there exist positive integers [math]q[/math] and [math]s[/math], such that:[math]p = q^2 … (6)[/math][math]16p^2+12p+3 = s^2 … (7)[/math]Now, by [math](7)[/math] we obtain:[math]16p^2+12p+3 = s^2 => 16p^2+12p+3-s^2 = 0 => [/math][math]p^2+(3/4)p+(3-s^2)/16 = 0 ... (8)[/math]Since [math]p[/math] is positive integer, the discriminant of [math](8)[/math] must be perfect square. The only possibility such that the discriminant of [math](8)[/math] to be equal to [math]0[/math], is when the trinomial of [math](8)[/math] with respect to [math]p[/math] has two equal real solutions. In our case, we also demand these two equal solutions [math](p[/math], [math]p)[/math] to be positive integers. This can happen, if and only if:[math](3-s^2)/16 = p^2 => 3-s^2 = 16p^2 => (4p)^2+s^2 = 3[/math],which is impossible since no sum of two squares is congruent to [math]3mod4[/math]. Hence, there exists positive integer [math]w[/math] such that:[math]144-64(3-s^2) = w^2 => 16[9-4(3-s^2)] = w^2 => [/math][math]9-4(3-s^2) = t^2 => 4s^2-t^2 = 3 => (2s)^2-t^2 = 3 … (9)[/math]where [math]t[/math] is positive integer.Now, the only possibility for a difference of two squares of positive integers [math]a[/math] and [math]b[/math] to be equal to [math]3[/math], is when [math]a = 2[/math] and [math]b = 1[/math]. Hence, if we let [math]a = 2s[/math] and [math]b = t[/math], we obtain:[math]2s = 2 => s = 1[/math] and [math]t = 1 … (10)[/math]By [math](10)[/math], equation [math](8)[/math] takes the form:[math]16p^2+12p+3 = s^2 => 16p^2+12p+3 = 1^2 => [/math][math]16p^2+12p+2 = 0 ... (11)[/math]Now, we see that equation [math](11)[/math] does not have a positive integer solution since its real solutions are [math]p = -1/2[/math] and [math]p = -1/4[/math]. Hence, when [math](p, 16p^2+12p+3) = 1[/math], the only posibility of [math](5)[/math] to have integer solution is when [math]p = 0[/math] and [math]m = 0[/math] or [math]16p^2+12p+3 = 0[/math] and [math]m = 0[/math]. The case [math]16p^2+12p+3 = 0[/math] must be rejected, since the discriminant of this trinomial is [math]-48 < 0[/math]. Therefore, since [math]p = 0[/math], it follows that [math]k = 0[/math] which means that:[math]b = 2k+1 => b = 2(0)+1 => b = 1[/math]In the case at which [math]gcd[p, 16p^2+12p+3] = 3[/math], we see that [math]p = 3r[/math], where [math]r[/math] is positive integer. Then, by [math](5)[/math] we have:[math]p(16p^2+12p+3) = m^2 => 3r(144r^2+36r+3) = m^2 => [/math][math]9r(48r^2+12r+1) = m^2 … (12)[/math]Now, we see that:[math]gcd[9r, (48r^2+12r+1)] = 1 … (13)[/math]Hence, by [math](13)[/math] it follows that there exist positive integers [math]c[/math] and [math]d[/math], such that:[math]9r = c^2 … (14)[/math][math]48r^2+12r+1 = d^2 … (15)[/math]By [math](14)[/math], we take:[math]r = (c^2)/9 … (16)[/math]By [math](16)[/math], equation [math](15)[/math] becomes:[math]48r^2+12r+1 = d^2 => 48((c^2)/9)^2+12(c^2)/9+1 = d^2 => [/math][math]48c^4+108c^2+81 = 81d^2 … (17)[/math]We set [math]y = 9d[/math] and equation [math](17)[/math] takes the form:[math]48c^4+108c^2+81 = y^2 … (18)[/math]Now, equation [math](18)[/math] can be modified to:[math]3(16c^4+36c^2+27) = y^2 … (19)[/math]By [math](19)[/math], we take:[math]16c^4+36c^2+27 = 3z^2 … (20)[/math]where [math]z[/math] is positive integer.Now, equation [math](20)[/math] can be modified to:[math]16c^4+36c^2+27 = z^2+z^2+z^2 … (21)[/math]If [math]c[/math] is odd, we observe that:[math]16c^4+36c^2+27 ≡ 7mod8 ... (22)[/math]By [math](22)[/math] and by Legendre’s three square theorem, equation [math](20)[/math] and hence equation [math](19)[/math], can not have positive integral solution.If [math]c[/math] is even, we observe that:[math]16c^4+36c^2+27 ≡ 3mod8 ... (23)[/math]Now if [math]c[/math] is even, there exists positive integer [math]t[/math], such that [math]c = 2t[/math], hence equation [math](20)[/math] takes the form:[math]16c^4+36c^2+27 = 3z^2 => 256t^4+144t^2+27 = 3z^2 … (24)[/math]By [math](24)[/math] we observe that [math]z[/math] must be odd. Hence, there exists positive integer [math]q[/math], such that [math]z = 2q+1[/math] and [math](24)[/math]becomes:[math]256t^4+144t^2+27 = 3(2q+1)^2 => 256t^4+144t^2+24 = 3(4q^2+4q) => [/math][math]64t^4+36t^2+6 = 3(q^2+q) => (8t^2)^2+(6t)^2 = 3(q-1)(q+2)[/math]which is not solvable in positive integers.Therefore, in any case, the only integral solutions of [math](20)[/math], are the trivial ones:[math](c = 0, z = -3), (c = 0, z = 3) [/math]Hence, the only integral solutions of [math](19)[/math] are:[math](c = 0, y = -9), (c = 0, y = 9)[/math]Therefore, since [math]c = 0[/math], by [math](14)[/math] we take:[math]r = (c^2)/9 => r = 0[/math]Now, since [math]p = 3r[/math], we take [math]p = 0[/math] and by [math]k = 2p[/math], we take [math]k = 0[/math], hence:[math]b = 2k+1 => b = 2(0)+1 => b = 1[/math]Now, let us assume that [math]b[/math] is even. Then, by [math](1)[/math], [math]a[/math] must be odd, hence there exist positive integers [math]p[/math] and [math]q[/math], such that:[math]b = 2p[/math] and [math]a = 2q+1 ... (25)[/math]By the above relations, equation [math](1)[/math] takes the form:[math](2p)^3 = (2q+1)^2+1 => 8p^3 = 4q^2+4q+1+1 => [/math][math]8p^3 = 4(q^2+q)+2 ... (26)[/math]Now, the LHS of [math](26)[/math] is congruent to [math]0mod4[/math], while the RHS of it is congruent to [math]2mod4[/math], which is a contradiction.Therefore, we have proved that the only solution of [math](1)[/math] is [math](a = 0[/math] , [math]b = 1)[/math].The solution of the above problem is dedicated to all who believe that we cannot solve advanced and difficult problems by elementary methods…!