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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.

Gottfried Helms
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math110
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  • 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 – math110 May 25 '17 at 12:51
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    inequality, where did you get the problem????? – Will Jagy May 25 '17 at 15:16
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    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
  • That is equivalent to showing that there are no non-trivial units of the form $m+n(\sqrt2+\sqrt3)$ in the ring $\Bbb{Z}[\sqrt2,\sqrt3]$. This is somewhat related but there was no proof that it describes the entire unit group. – Jyrki Lahtonen May 25 '17 at 18:50
  • Following @Gottfried Helms idea, I just posted a proof here. – René Gy May 07 '18 at 21:37
  • @JyrkiLahtonen - I just updated my answer matching your first comment; but just using that equation ($s=m^2+n^2$) which you'd dismissed as uninteresting gets the goal. – Gottfried Helms Dec 06 '20 at 17:57
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    @GottfriedHelms The reason I called that "uninteresting" is that the Pythagorean stuff made it automatic that $3k^2+1$ is a perfect square (and plugging in $s=m^2+n^2$ yields the same equation). Happy to be proven wrong :-) – Jyrki Lahtonen Dec 06 '20 at 18:46

3 Answers3

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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.

GNUSupporter 8964民主女神 地下教會
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Angus Chung
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  • 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
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    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? – CryoDrakon Apr 28 '18 at 13:56
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    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
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    @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
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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. \ $

Somos
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Update: Just found the solution; see end of text (taken from my other answer )

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.

Update: Just found the solution; it is arriving at the same observation which was pointed at by @JyrkiLakhonen in his first comment.

Using $3x_h^2 +1 = t_h^2$ and the assumption that there exist an entry $y_i$ in $2y_i^2+1=u_i^2$ such that $y_i=x_h$ and thus $2x_h^2+1=u_i^2$ we can subtract the two equations and get $x_h^2 = t_h^2-u_i^2$ .

This is a pythagorean triple, for which a parametric solution exists in natural numbers$(n,m)>0$. From this we have that $x_h = m^2-n^2$ and $t_h=m^2+n^2$ as Jyrki already mentioned (but dismissed as uninteresting). This says, that the parity of $x_h$ and $t_h$ must be equal to have a solution.

But if we look at the sequences $x_h$ and $t_h$ we see that the parity is always opposite. From this follows that there is no integer solution possible. (The fact that the parity must always be opposite is easy to prove, I think).

Gottfried Helms
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