TITLE: Let $k$ be a postive integer number . Then $2k^2+1$ and $3k^2+1$ cannot both be square numbers.
QUESTION [19 upvotes]: Let $k$ be a postive integer number . Then $2k^2+1$ and $3k^2+1$ cannot both be square numbers.

I tried to prove this by supposing one of them is a square number and by substituting the corresponding $k$ value. But I failed to prove it.

REPLY [3 votes]: The two Diophantine equations are equivalent to an elliptic curve with Cremona label "96b1" or LMFDB label 96.a3.
The equation of the curve is $\ E: y^2 = x^3 - x^2 -2x = x(x+1)(x-2). \ $ The rational points generate a Klein four-group with points $\ (0,0), (2,0), (-1,0) \ $ along with the point at infinity. There is an algebraic point $\ (3,\sqrt{12}) \ $ of infinite order. Each point on this elliptic curve corresponds to a solution of
the system of homogeneous quadratic equations
$$ X^2 - Y^2 = 1 W^2, \quad Y^2 - Z^2 = 2W^2, \quad X^2 - Z^2 = 3W^2 $$
 which comes from a solution of the system of equations
 $$ v^2 - u^2 = 1k^2, \quad u^2 - 1 = 2k^2, \quad v^2 - 1 = 3k^2 $$ by
dehomogenizing as $\ k = W/Z, \quad u = Y/Z, \quad v = X/Z. \ $ The only rational solutions are $\ k = 0,  \quad u = \pm 1,  \quad v = \pm 1. $ An infinite family of algebraic solutions is generated by the particular solution $\ W^2 = 1,  \quad Z^2 = 1,  \quad  Y^2 = 3,  \quad X^2 = 4. \ $