# Let $k$ be a postive integer number . Then $2k^2+1$ and $3k^2+1$ cannot both be square numbers.

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.

• What do you mean by "cannot all be square numbers"? The same as here? – Dietrich Burde May 25 '17 at 12:46
• @DietrichBurde I think he means "it is not possible for both $2k^2+1$ and $3k^2+1$ to be perfect squares". – Parcly Taxel May 25 '17 at 12:47
• @DietrichBurde,yes,That's right – inequality May 25 '17 at 12:51
• inequality, where did you get the problem????? – Will Jagy May 25 '17 at 15:16
• Assuming that $2k^2+1=r^2$ and $3k^2+1=s^2$ for some positive integers $k<r<s$ then $(k,r,s)$ is a coprime Pythagorean triple. By the well known parametrization of those (and the fact that necessarily $r$ is odd) we see that there exists natural numbers $m,n$ such that $k=2mn$ and $r=m^2-n^2$ ($s=m^2+n^2$ becomes uninteresting). The equation $2k^2+1=r^2$ then translates to $$1=m^4-10m^2n^2+n^4.$$ It does look like that $\{m,n\}=\{0,1\}$ is the only solution of that (leading to $k=0$), but I don't know how to prove that, One of $m,n$ must be divisible by five, but no cigar :-( – Jyrki Lahtonen May 25 '17 at 17:42

This basically continues from Jyrki Lahtonen's idea in the comment.

Suppose not. The triple $(2k^2+1, k^2, 3k^2+1)$ would be squares of Pythagorean triple, and clearly they are relatively prime (note: NOT mutually coprime). Thus there are $m$, $n$ coprime integers such that

\begin{align} m^2-n^2 &= 2k^2+1 \\ 2mn &= k^2 \\ m^2+n^2 &= 3k^2+1 \end{align}

(Note that among $2k^2+1$ and $k^2$, the latter is clearly the even one.)

There are a lot of ways from here. For example, $2n^2 = k^2$, a contradiction.

• Please typeset the equation in the middle with MathJax. – GNUSupporter 8964民主女神 地下教會 Apr 28 '18 at 13:20
• Maybe I am mistaken, but you said let it be true and then we have $a^2+k^2=b^2$, where $a^2=2k^2+1$, and $b^2=3k^2+1$. Now we know that $a=m^2-n^2, k=2mn, b=m^2+n^2$. Don't you then miss a sqareroot on the right side? – HeatTheIce Apr 28 '18 at 13:56
• Comments can easily and irremediably be deleted. It's a good idea to not "continue" a comment and instead make a self-contained answer. – Najib Idrissi Apr 28 '18 at 15:30
• @HeatTheIce Right! If $2k^2+1=A^2$ and $3k^2+1=B^2$ then $(k,A,B)$ is a primitive Pyth. triplet so $\{k^2, 2k^2+1\}=\{4mn, (m^2-n^2)\}$ and $3k^2+1=m^2+n^2.$.... The squares of a Pyth, triplet cannot also be a Pyth, triplet because $a^4+b^4=c^4$ has no solution in positive integers. – DanielWainfleet Apr 28 '18 at 18:55
• Hmm, I think there's something basically wrong. We know, that in $A^2 = 2k^2+1$ there is one solution $A=3,k=2$. So $2mn=k^2=4$ and $mn=2$, and because $m^2-n^2=9$ it must be that $m>n$, so by $mn=2$ we must have $m=2$ and $n=1$. But $2^2-1^2=3$ and not $9$ . So for this case the first two equations lead to a contradiction. – Gottfried Helms Jul 20 '18 at 12:12

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. \$

Not yet an answer, but another ansatz, using transfer-matrices:

A: Let's look at the solutions $3x^2+1 = t^2$ We find $x_0=0$ and $x_1=1$ are the first solutions giving $t_0=1$ and $t_1=2$.
Heuristically the next solutions $x_h=\{0,1,4,15,56,...\}$ and $t_h=\{1,2,7,26,97,...\}$ can be generated by $$\begin{bmatrix}x_0&x_1 \\ t_0&t_1 \end{bmatrix}\cdot M_3 ^h = \begin{bmatrix}x_h&x_{h+1} \\ t_h&t_{h+1} \end{bmatrix}$$ with a transfermatrix $$M_3=\begin{bmatrix} 0&-1 \\1&4 \end{bmatrix}$$

B: Let's next look at the solutions $2y^2+1 = u^2$ We find $y_0=0$ and $y_1=2$ are the first solutions giving $u_0=1$ and $u_1=3$.
Heuristically the next solutions $y_i=\{0,2,12,70,408,...\}$ and $u_i=\{1,3,17,99,577,...\}$ can be generated by $$\begin{bmatrix}y_0&y_1 \\ u_0&u_1 \end{bmatrix}\cdot M_2 ^i = \begin{bmatrix}y_h&y_{i+1} \\ u_h&u_{i+1} \end{bmatrix}$$ with a transfermatrix $$M_2=\begin{bmatrix} 0&-1 \\1&6 \end{bmatrix}$$

C: Now the problem is to show, that $x_h \ne y_i$ for all $h,i>0$. It looks like a combined Pell-problem whose individual solutions are also computable by simple recursions resp. powers of the associated transfermatrices.

Here I tried this by eigendecomposition of $M_3$ and $M_2$ (which leads to Binet-like formulae as they occur for the recursion in the Fibonacci-sequence) but did not yet arrive at a conclusion...

It is interesting to look at the prime-factorizations of the sequences $x_h$ and $y_i$ The primefactors occur periodically with the indexes and a lot of (small) cases can be excluded as possible solutions just by cycle-lengthes of combinations of small primefactors.