Consecutive partitions of positive integers

[Edited to include the connection with Euler's theorem that there's no nonconstant $4$-term arithmetic progression of squares; briefly, since $1$ and $4(N^2+N)+1 = (2N+1)^2$ are always squares, $(1,b,c,(2N+1)^2)$ is such a $4$-term progression, so $b=c=1$ and $N^2+N=0$ for any rational solution. It seems that the same deals with all $k \geq 3$.]

The Diophantine equations $$ B^2 = b = \frac43(N^2+N) + 1 $$ $$ C^2 = c = \frac83(N^2+N) + 1 $$ have no simultaneous solution in integers other than the eight trivial solutions with $N=0,-1$, $B = \pm 1$, $C = \pm 1$. Such a result can be tricky to prove in general, but here we are somewhat lucky in that these are the only rational solutions, and moreover $N=0$ and $N=-1$ are the only rational numbers for which the product $bc$ is a square. This can be proved using Fermat's "descent" method; the calculations, though elementary, may be somewhat laborious to carry out by hand, but happily this is no longer necessary thanks to existing tables and software.

In general, if $P,Q$ are quadratic polynomials such that $PQ$ has four distinct factors then the simultaneous Diophantine equations $B^2 = P(N)$, $C^2 = Q(N)$ define an elliptic curve. A fundamental 1929 theorem of Siegel assures that such a curve has only finitely many integral points. The proof is "ineffective" and does not in general give an algorithm guaranteed to determine all solutions. Later theoretical and computational advances do provide such an algorithm, which is feasible at least for $P,Q$ with small coefficients; but the resulting proof is very far from elementary, and it can be hard to predict in any given case how hard it is to give an elementary proof.

In the present case, though, the elliptic curve already has finitely many rational points, as does the "isogenous" curve $A^2 = P(N) Q(N)$. We bring this curve $$ A^2 = \left( \frac43(N^2+N) + 1 \right) \left( \frac83(N^2+N) + 1 \right) $$ into standard Weierstrass form in the usual way starting from the rational point $(N,A) = (0,1)$: the Taylor expansion of $A$ about $N=0$ starts $1 + 2N + O(N^2)$, so for $N\neq 0$ we can write $$ A = 1 + 2N + rN^2 $$ for some rational $r$, and divide the resulting equation by $N^2$ to get $$ (9r^2-32)N^2 + (36r-64)N + (18r-32) = 0. $$ This equation is quadratic in $N$, and thus has rational roots iff its discriminant w.r.t. $N$ is a square. That discriminant factors as $-72(r-2)r(9r-16)$; taking $r=-2x/9$ and removing a factor $(8/3)^2$ we find that $$ y^2 = x (x+8) (x+9) = x^3 + 17 x^2 + 72 x $$ for some rational $x,y$. This elliptic curve turns out to have conductor $24$, so it already appears in Tingley's "Antwerp Tables" of modular elliptic curves of conductor at most $200$; it turns out to have label 24C here. We find that it has rank zero, and only four rational points, which must be the point at infinity and the three "$2$-torsion points" with $y=0$. Alternatively, we can input [0,17,0,72,0] to Cremona's program mwrank to find that the curve has rank zero, and then find its torsion points with gp. Either way, we finish by retracing our steps to find that each of $r=0,2,16/9$ corresponds to one of the known solutions with $N=0$ or $N=-1$ (in each case a double root of the quadratic in $N$), so we are done.

[Thanks to Will Jagy for calling my attention to this question.]

Added later: It turns out that this $2$-descent calculation long predates the Antwerp tables: it is essentially equivalent to Euler's proof of his theorem that there is no nonconstant $4$-term arithmetic progression of squares. (See for instance Keith Conrad's exposition, which states that Euler proved the result in 1780, answering a question "first raised by Fermat in 1640".) Indeed if $b$ and $c$ are squares then $$ 1, \ \frac43(N^2+N)+1, \ \frac83(N^2+N)+1, \ 4(N^2+N)+1 $$ is such a progression (the last term being $(2N+1)^2$, so it is constant by Euler, whence $b=c=1$ and we are done.

jamaicanworm writes in a comment that for general $k \geq 3$ the problem is whether there exists a positive integer $N$ such that $c_i := \frac{4(k-i)}{k}(N^2+N) + 1$ is a square for each $i=1,2,\ldots,k-1$. These $c_i$ form an arithmetic progression, and extending it by one term in each direction again yields the squares $c_k = 1$ and $c_0 = (2N+1)^2$. Hence we have an arithmetic progression of $k+1$ squares, and again Euler's theorem shows that even if we allow $N$ to be rational the only examples are the trivial ones with $N^2+N=0$.

I can help you anwer the first question: This has something to do with solutions to Pell's equation. I let mathematica solve it, and it gave the following solution: $$ \frac{1}{4} \left(\sqrt{2} \left(3+2 \sqrt{2}\right)^{c_1}-\left(3+2 \sqrt{2}\right)^{c_1}-\sqrt{2} \left(3-2 \sqrt{2}\right)^{c_1}-\left(3-2 \sqrt{2}\right)^{c_1}-2\right) $$ where $c_1$ is an integer. The system of equations for question two can't be solved by mathematica, but i do not se a simple proof for that right now. Whether or not a $k$-partition exists for an integer $k$ is not an easy question, but maybe you can express a system of equations for arbitrary $k$ and show that it can't have solutions.

EDIT The command i used wasSolve[2n^2+2n+1==k^2&&n>0&&k>0,{n,k},Integers]. This way, it can solve this kind of formula's. An online solver is available here. This returns a recurrence relation to generate the consecutive values of $n$ given by the formula above. How to solve a general linear recurrence relation can be found here.

EDIT 2 I just realized the following: $3$-consecutive equal sums can only occur when the first two are equal, and thus, the largerst element of the middle sum has to be a value of $N$ which comes from the formula above and sattisfies the $2$-consecutive sum criterium. I will try to proof that $k$-consecutive sums do not exist for $k\geq 3$.