Let $k$ be a natural number . Then $3k+1$ , $4k+1$ and $6k+1$ cannot all be square numbers.

Let $k$ be a natural number . Then $3k+1$ , $4k+1$ and $6k+1$ cannot all 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.

Solutions Collecting From Web of "Let $k$ be a natural number . Then $3k+1$ , $4k+1$ and $6k+1$ cannot all be square numbers."

Complete answer from the literature:

I found a list of solved Pell systems in Szalay, Appendix 4, page 84.

Kiran Kedlaya published Solving Constrained Pell Equations in Mathematics of Computation, Volume 67, April 1998, pages 833-842.

As an example of his method, on page 840 he does the harder part of the Lucas problem (bottom page 238), namely

Solving $n+1 = 2 x^2,$ $2n+1 = 3 y^2,$ $n=z^2$

with the conclusion

Possible values of $n:$ $\{1\}$

This fits the problem above by taking $x^2 = 3k+1,$ $y^2 = 4k+1,$ $z^2 = 6k+1$ which gives the system
$$ 2 x^2 – z^2 = 1,$$
$$ 3 y^2 – 2 z^2 = 1, $$
with the conclusion that we can only have $z^2 = 1,$ so $k=0.$

Kedlaya refers to his own solution as elementary, references bottom of page 838 to top of page 839. He mentions other articles with elementary methods; I think this is a case where “elementary” is in the eye of the beholder.

I think this will work, but I don’t quite have time today to fill in all details.

If $3k+1=x^2,4k+1=y^2,6k+1=z^2$, then $2x^2+z^2=3y^2$. Letting $a=x/y,b=z/y$, this gives $2a^2+b^2=3$. We can parametrize the whole space of solutions like one usually does for Pythagoream triples: $(1,1)$ is an obvious point on the curve. Now let $t$ be any real and let $Y=t(X-1)+1$ be a line through it. It intersects the curve $2a^2+b^2=3$ at point $(1,1)$ and other point $(a,b)$ satisfying $2a^2+b^2=3$ and $b=t(a-1)+1$. Substituting, we get $a^2(2+t^2)+a(2t-2t^2)+(t^2-2t-2)=0$. If we interpret this as a quadratic in $a$, $1$ is one solution, so by Viete’s formulas, the other is $a=(t^2-2t-2)/(t^2+2)$. Then $b=(-t^2-4t+2)/(t^2+2)$. Now we look at $x/z=a/b=(t^2-2t-2)/(-t^2-4t+2)$. Letting $t=n/m$ be rational, this gives $x/z=(n^2-2nm-2m^2)/(-n^2-4nm+2m^2)$.

Clearly $x,z$ are relatively prime, since $2x^2-z^2=1$. I claim $n^2-2nm-2m^2,-n^2-4nm+2m^2$ have $\gcd$ dividing $6$. This is because their sum is $6nm$, and if $p\mid n$ and $p\mid n^2-2nm-2m^2$, then $p\mid m$ and vice versa.

(to be continued)

Assuming the contrary, set $x^2 = 3k+1$, $y^2 = 4k+1$ and $z^2 = 6k +1$

Now, has anyone tried an approach with Pell’s equations?

Playing around with these three squares we get

  • A: $4x^2 -3y^2 = 1$
    which is equivalent to $u^2 – 3y^2 =1$, for $u=2x$
  • B: $3y^2 – 2z^2 = 1$

and

  • C: $z^2 – 2x^2 = -1$

The idea is to show that there is no integer $x,y,z$ satisfying A,B and C at once. All three equations have been already studied extensively but the messy nature of the solutions of B are making it hard for me to get a contradiction.

$\def\Q{\mathbf{Q}}$
$\def\Qbar{\overline{\Q}}$
$\def\Z{\mathbf{Z}}$
$\def\F{\mathbf{F}}$

I want to try to explain why there won’t be any slick elementary argument in this case. The usual competition techniques involve a tricky use of congruences. A more advanced trick is to use infinite descent. However, in the end, this problem boils down to finding all integral points on an elliptic curve with positive rank (the curve $Y^2 = X^3 – 36 X$). There are some pretty standard techniques to do this – but none of them are elementary. The reason why elementary arguments (such as congruences) are doomed to fail is related to the fact that there are many solutions to this problem with $k \in \Q$.

Any odd square is $1 \mod 8$. Hence $k$ is even, and so we need to find $k$ such that
$6k+1$, $8k+1$, and $12k + 1$ are squares.

As others have observed, can write down equations for this problem as $a^2 = 6k+1, b^2 = 8k+1$, and $c^2 = 12k+1$. This leads to the sytem equations
$$C: 3 b^2 – 4 a^2 = -d^2, \quad c^2 – 2 a^2 = -d^2,$$
where $d = 1$. Here I introduced the variable $d$ to turn the affine equation into a projective one.
The intersection of two quadrics in $\mathbf{P}^3$ has genus one, and so this equation is an elliptic curve (for some choice of base point, note that $[1;1;1;1] \in C(\Q)$.)
It is a general theorem (of Siegel) that, given an elliptic curve has only finitely
many integral points. On the other hand, most elementary solutions to these sort of problems
typically prove
the stronger claim that $E(\mathbf{Q})$, the set of rational points, is also finite. Proofs of this kind of
statement were first found by Fermat, and use the method of infinite descent.
For example, Fermat proved that there are no integral solutions to $x^4 + y^4 = z^4$ by considering
the equation $x^4 + y^4 = z^2$ in rational numbers. After scaling, one can let $y = 1$, and then $z^2 = x^4 + 1$ is an affine open inside an elliptic curve.
Unfortunately, as we shall see, this is not one of those examples.

For those who like to think of elliptic curves in terms of Weierstrass equations, it’s easy to write down an elliptic curve $E$ isogenous to $C$ of
the form $y^2$ is a cubic. Namely, if $6k + 1$, $8k+1$, and $12k+1$ are all squares, then so is their product, hence we also get a solution
in integers (and hence in rational numbers) to the equation

$$E: z^2 = (6x+1)(8x+1)(12x+1),$$

Note that by replacing $k$ with $2k$ (which makes no difference to rational solutions), it makes the coefficient of $x^3$ equal to a square ($576 = 24^2$). And now, letting $z = 24y$, we can also write as

$$E: y^2 = (x + 1/6)(x + 1/8)(x + 1/12),$$

and the point $\infty$ at infinity is a rational point.
There is a degree $2$ isogeny $C \rightarrow E$ given by
$$(a,b,c) \mapsto \left(\frac{a – 1}{6},\frac{abc}{24} \right).$$
There will also be a dual isogeny $E \rightarrow C$ of degree $2$, but it will be a little messy to write down.

Descent: The main elementary way to find rational points on an elliptic curve $E$ is
to do a $2$-descent. In the case
when all the $2$-torsion is rational, this is pretty concrete. Explicitly, there is an injective map:
$$E(\Q) \rightarrow \Q^{\times}/\Q^{\times 2} \times \Q^{\times}/\Q^{\times 2} \times \Q^{\times}/\Q^{\times 2}$$
which sends a point $P = [x,y]$ to the class:
$$(x + 1/6, x + 1/8,x + 1/12).$$
Note that this class lies inside the subgroup $(a,b,c) \subset (\Q^{\times}/\Q^{\times 2})^3$
with $abc = 1$, so we can really just take the first two entries in
$(\Q^{\times}/\Q^{\times 2})^2$.
The key fact about this map is that the kernel is precisely $2 E(\Q)$, so the induced
map from $E(\Q)/2 E(\Q)$ is injective.
A slight subtlety — this doesn’t quite make sense
for $2$-torsion points, because one of the entries is zero. But you can work out the
correct image by noting that it should land inside the diagonal. Hence
$$\infty \mapsto (1,1,1),$$
$$e_1 = [-1/6,0] \mapsto (*, -1/24,-1/12) \equiv (2,-6,-3) \in (\Q^{\times}/\Q^{\times 2})^3,$$
$$e_2 = [-1/8,0] \mapsto (1/24,*,1/24) \equiv (6,-1,-6) \in (\Q^{\times}/\Q^{\times 2})^3,$$
$$e_1 + e_2 = [-1/12,0] \mapsto (1/12,1/24,*) \equiv (3,6,2) \in (\Q^{\times}/\Q^{\times 2})^3.$$

On the other hand, $E$ also has an obvious rational point $P = [0,1]$. We find that
$$P \mapsto (6,2,3) \in (\Q^{\times}/\Q^{\times 2})^3,$$
$$P + e_1 \mapsto (3,-3,1) \in (\Q^{\times}/\Q^{\times 2})^3, $$
$$P + e_2 \mapsto (1,-2,-2) \in (\Q^{\times}/\Q^{\times 2})^3,$$
$$P + e_3 \mapsto (2,3,6) \in (\Q^{\times}/\Q^{\times 2})^3.$$

Let’s now make the following observation, using the fact that $\mathbf{Q}$ has unique factorization.
If $(x + 1/6)$, $(x + 1/8)$, and $(x + 1/12)$ have a square product, then they are all squares up to possible common factors. (Any power of $p > 3$ in the numerator will have to be a square as well by an elementary computation). However, any common factor must divide
$$\Delta = (1/6 – 1/8)^2(1/6 – 1/12)^2 (1/8 – 1/12)^2 = 2^{-16} \cdot 3^{-6}.$$
This implies that the image of $E(\mathbf{Q})/2 E(\mathbf{Q})$ lands in the smaller group
$$(S^{\times}/S^{\times 2 } )^2,$$
Where $S \subset \mathbf{Q}^{\times}$ is generated by $2$, $3$, and $-1$.
This is a group of order $8 \times 8 = 64$. We have computed a subgroup of order $8$ that lies in the image.
If we are lucky, we can show that none of the other $56$ points lie in the image for local reasons, and this would imply that $E(\mathbf{Q})/2 E(\mathbf{Q}) = (\mathbf{Z}/2 \mathbf{Z})^3$.
We don’t need to check all $56$ cases, because the image is a group. And most of them are easy. Namely, we write:
$$x + \frac{1}{6} = a u^2, x + \frac{1}{8} = b v^2, x + \frac{1}{12} = a b w^2,$$
or
$$a u^2 – b v^2 = \frac{1}{24}, \qquad a u^2 – a b w^2 = \frac{1}{12},$$
where $(a,b) \in (S^{\times}/S^{\times 2 } )^2$.
We can immediately rule out many examples: If $b > 0$ and $a < 0$ the first equation has
no solutions. If $a < 0$ and $b < 0$, the second equation has no solutions. Hence $a > 0$.
This leaves us with $32$ possibilities. Considering the equation $2$-adically rules out half of the remaining possibilities, and $3$-adic considerations reduce us by half again to end up with $8$.
Thus we have shown:

$$E(\mathbf{Q}) \simeq \mathbf{Z} \oplus (\mathbf{Z}/2 \mathbf{Z})^2,$$
and $P = [0,1]$ is a generator. (Computing the torsion subgroup is easy:
note that $|E(\F_{11})| = 12$ and $E(\F_{5}) = 8$, so the torsion subgroup has order dividing $4$.)

So we have now found (in some sense) all rational points on $E$. However, the answer is a little unexpected, because we had hoped to show that the original equation had no solutions by showing
that there were no rational solutions. But there are many rational solutions!

Let’s think of solutions to $E(\mathbf{Q})$ that actually come from $C(\mathbf{Q})$.
We want $6x+1$, $8x +1$, and $12x+1$ to all be squares. Hence, we want the image
of the point in $E$ in $(\mathbf{Q}^{\times}/\mathbf{Q}^{\times 2}$ to be
$$(1/6,1/8,1/12) = (6,2,3).$$
Clearly $P$ is such a point, and this corresponds to $x = 0$ or $k = 0$.
Any other point which maps to this must be equal to $P$ in $E(\mathbf{Q})/2 E(\mathbf{Q})$,
and thus $P \mod 2 E(\mathbf{Q})$. Hence we deduce:

Rational points on $C(\mathbf{Q})$ correspond to points on $E(\mathbf{Q})$ of the form
$2Q + P$, or equivalently, $n P$ for $n$ odd.

For example,
$$3P = \left[ \frac{-140}{2209},\frac{-28083}{103823} \right],$$
and we correspondingly see that
$$3 \left(\frac{-280}{2209}\right) + 1 = \left(\frac{37}{47}\right)^2, 4 \left(\frac{-280}{2209}\right) + 1 = \left(\frac{33}{47}\right)^2, 6 \left(\frac{-280}{2209}\right) + 1 = \left(\frac{23}{47}\right)^2$$
Taking $5P$ gives
$$3 \left(\frac{369572280}{537451489}\right) + 1 = \left(\frac{40573}{23183} \right)^2,
4 \left(\frac{369572280}{537451489}\right) + 1 = \left(\frac{44897}{23183}\right)^2, 6 \left(\frac{369572280}{537451489}\right) + 1 = \left(\frac{52487}{23183}\right)^2,$$
and so on.

But what does this say about our original question? A weaker question than asking
for solutions with integral $x$ is to ask for solutions with $72x \in \mathbf{Z}$. Then we can
make the transformation
$$X = \frac{x}{144} – \frac{1}{8}, \quad Y = \frac{y}{1728},$$
to get the equation
$$D: Y^2 = X^3 – 6^2 X.$$
This is isomorphic to $E$ over $\Q$, but it is a different integral model. This is a familiar curve; it is the congruent
number curve for $d = 6$. There is an integral solution related to the fact that $(3,4,5)$ is a
right angled triangle with area $6$.
But there are other integral solutions as well, namely:
$$[0,0], [\pm 6, 0], [-3, \pm 9], [-2, \pm 8], [12, \pm 36],
[18, \pm 72], [294, \pm 5040].$$
This is, in fact, the complete list. If you like, you can confirm this in magma as follows:

$$\begin{aligned}
& \ \texttt{> E := EllipticCurve([0,0,0,-36,0]);} \\
& \
\texttt{> IntegralPoints(E);} \\
& \ \texttt{
[ (-6 : 0 : 1), (-3 : -9 : 1), (-2 : 8 : 1), (0 : 0 : 1),} \\
& \ \texttt{ (6 : 0 : 1), (12 : -36
: 1), (18 : 72 : 1), (294 : -5040 : 1) ]
} \end{aligned}$$

The observatio that $E$ has many rational points and lots of $S$-integral
points for $S \in \{2,3\}$ suggests that there won’t be a quick
solution to the original problem.
There are fairly standard methods to (certifiably) compute all the integral
points of an elliptic curve $D$ given the Mordell-Weil group (which
$\texttt{magma}$ uses), but they are not particularly
elementary.

This is not a complete answer, but perhaps it will help someone. If $3k = a^2-1$ and $6k = c^2-1$, then $c^2-2a^2 = -1$. It follows that the possibilities for $c$ and $a$ are the odd convergents to the continued fraction for $\sqrt{2}$ (thus $k=8$, corresponding to $c=7$ and $a=5$, is the first such value).

[update] I’ve arrived at a short formula of $\sinh()$, $\operatorname{asinh}()$ which must be proved to arrive at integers only; accepting an asymptotic version the formula becomes even simpler and reduces to logarithms. However, I do not have the knowledge to do that proof, so this answer stays unfinished. I do not know, whether this is of interest for you, but if it is I can add the detailed derivation into this post.
The formula in its most simple (but asymptotic form) works as follow.
Assume some $c^2$ from table “a_c” (see bottom of post) then use $h$ for its row-index (beginning at zero). If the same value $c^2$ occurs also in table “_bc” then it has there another row-index, call it $i$. If $h-i$ is integer, then we have a solution.

It is now possible to find Binet-like formulae for $c_h$ in table “a_c” as well as for $c_i$ in table “_bc”.
Let $q=3 + \sqrt 8$ for table “a_c” then $c_h = {q^h \over 1-1/q} + {q^{-h}\over 1-q}$.
Let $r=5 + \sqrt{24}$ for table “_bc” then $c_i$ can be computed by the same formula where $q$ is replaced by $r$ and the index $h$ by the index $i$.

Asymptotically we get then for some initial rowindex $h$ with which we select one $c_h^2$ the index $i$ at which that value would occur in table “_bc”.
Because of the Binet-form of the formulae, that indexes can (and shall be) fractional, and we want to find such $h$, $c_h^2$ for which $i$ is integer and thus a true index in the table “_bc”. The asymptotical formula is
$$ i = \log_r \left(q^h {r-1\over q-1}{q\over r} \right) $$
where $\log_r(x) = \log(x)/\log(r)$ or
$$ r^i {r \over r-1 } = q^h {q\over q-1} $$

For an example, if you use the fractional rowindex $h=2.55462$ you get a $c_h \approx 109.003$ (which is not in table “a_c” but between $41$ ad $239$ ) but this gives the index $i \approx 2$ with a value $c_i \approx 109$ in table “_bc”. The difference $h-i$ is of course non-integer, so this shows, that $c=109$ does not provide a solution.

As I said in the operning: because I can not prove that integer-property this is still not a complete answer, but might be helpful to proceed on your own…
[end update]

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I like to approach this problem by a somehow “constructive” approach. I did not yet find the final proof but a very promising path.

First idea: display a table of index $k$ and look, at which $k$ the values $3k+1$, $4k+1$, $6k+1$ are squares. Here is a short extract from that table:

  k|3k+1  4k+1  6k+1
   |?=a²  ?=b²  ?=c²
---------------------
   0   1    1    1
   1   4    .    .
   2   .    9    .
   3   .    .    .
   4   .    .   25
   5  16    .    .
   6   .   25    .
   7   .    .    .
   8  25    .   49
   9   .    .    .
  10   .    .    .
  11   .    .    .
  12   .   49    .
  13   .    .    .
  14   .    .    .
  15   .    .    .
  16  49    .    .
  17   .    .    .
  18   .    .    .
  19   .    .    .
  20   .   81  121
  21  64    .    .

First we find, that there are many rows where no square occurs; that rows should be deleted and that according $k$’s are to be ignored. But then we see, that in the first column $a^2$ each third square is missing, in the column $b^2$ only the squares of $b=1$ or $b=-1 \pmod 4$ are present, and in column $c^2$ only squares of $c=1$ and $c=-1 \pmod 6$ are present. In short: in each column occur only two sequences: one sequence where $a= 1 \pmod 3$ and another one where $a=-1 \pmod 3$, then in column $b^2$ where $b=1 \pmod 4$ and where $b=-1 \pmod 4$ and similarly in the $c^2$ column.


The next step is to find structures of the index $k$ (or of the unsquared values $a,b,c$) where squares in at least 2 columns $a^2,b^2,c^2$ occur; they are rare and one example is at $k=20$ with $b^2=81,c^2=121$ or $b=9,c=11$.

A first analytical ansatz is thus the following table where the free parameters $j,m,n$ define the occuring squares (the counting begins at zero):
$$ \begin{array} {rll|lll}
\text{col}& \text{sequence } 1& \text{sequence }2 & \text{composition of the}&\text{occuring }k \\
a^2: & a^2=(3j+1)^2 & A^2 =(3j+2)^2 & k=j(3j+2) & K=(j+1)(3j+1)\\
b^2: & b^2=(4m+1)^2 & B^2 =(4m+3)^2 & k=2m(2m+1) & K=2(m+1)(2m+1)\\
c^2: & c^2=(6n+1)^2 & C^2 =(6n+5)^2 & k=2n(3n+1) & K=2(n+1)(3n+2)\\
\end{array}$$
(Here I also used small and capital letters to discern between the two (interleaved) sequences for each column).

So to find a row $k$ where in $a$ and $b$ a square occurs simultaneously, the $k$ from an $a$ is shown in the first row: $k=j(3j+2)$ and the same $k$ from a $b$ is then found when also $k=2m(2m+1)$ and thus all solutions where the two first sequences of column $a^2$ and $b^2$ match are defined by the solutions of $j(3j+2)=2m(2m+1)$. Of course we see immediately that this can only happen for even $j$ and thus for $a$ of the form $a=6j’+1$. And so on…

I’ve tried a bit to do the relevant analyzes but have not yet analytically derived (or confirmed) the formula for the cases where at least two columns match with squares on one $k$’ th row; I think, I can provides this later.


But that ansatz gave the idea for an impressive heuristic table with a not-too-difficult structure, which show the path, how one should proceed analytically from the previous steps. Here is the table of only that rows, where squares occur at least in two columns simultanously. Furthermore I’ve ordered and separated that table in three partial tables and find elegant recursive formulae for the columns which strongly suggest, that there are no solutions where all three columns have a square on then same row, because the two partial sequences of each columns are completely listed in the three tables and by the extremely clear structure there seem to be no room in the scheme for a “surprising” match of all tables in one special row.

Here are three tables for the matches of columns $a,c$, $b,c$ and $a,b$ The only threefold match $a,b,c$ is $k=0,a^2=b^2=c^2=1$

table a_c:
picture
The recursion for the $k$-entries is $k_0=0,k_1=8$, $k_{i+2}=34 k_{i+1} – k_i + 8$

table _bc:
picture
The recursion for the $k$-entries is $k_0=0,k_1=20$, $k_{i+2}=98 k_{i+1} – k_i + 20 $

table ab_:
picture
The recursion for the $k$-entries is $k_0=0,k_1=56$, $k_{i+2}=194 k_{i+1} – k_i + 56 $


The columns $k$ in each table have the first nonzero value as common factor. If we build the lcm() of the three common factors $8,20,56$ we get the number $280$ (which @Lucian had already pointed at) which must thus be a factor of the $k$-value for any perfect triplett solution.

I think this structures help to find an elementary access onto the deeper (and required) analytical considerations.

Edit. There must be an error in the following answer, I cannot locate it yet, will leave it as is.

This is a reduction of the problem to another problem that I could not solve.

Are there positive integers $t,i,j$ with $j>i$ such that:

$\displaystyle \frac{t(t+1)}2=\frac{2i(j-i)j(j+i)}3$ ?

I do not know if the above problem is any easier or more elementary.
The reduction to this problem is elementary (using Pythagorean triples), as follows.

Let $x^2 = 3k+1$, $y^2 = 4k+1$, $z^2 = 6k+1$ (as in @WillJagy ‘s answer: I do not know if the other answers already might contain something that is equivalent to my attempt).

Then $(x^2)^2=(3k+1)^2=(3k)^2+6k+1=(3k)^2+z^2$, hence $x^2=m^2+n^2$, $3k=2m n$, $z=m^2-n^2$ (for some integers $m>n>0$), where I used that $z$ is odd, $k$ is even.

$1=x^2-3k=m^2+n^2-2m n=(m-n)^2$,
hence $m=n+1$, $x^2=2n^2+2n+1$, $z=2n+1$, $3k=2(n+1)n=2n^2+2n$.

Since $x^2=m^2+n^2$, we have $x=j^2+i^2$, and
either $m=2ij, n= j^2-i^2$, or $m= j^2-i^2, n=2ij$ for some integers $j>i>0$
(the case $i=0$ resulting only in the solution $(x,y,z)=(1,1,1)$).
Either way, $mn=(n+1)n=2ij(j^2-i^2)=2i(j-i)j(j+i)$.

$\displaystyle z^2-y^2=2k=\frac{4(n+1)n}3$, hence (using that $z=2n+1$)
$\displaystyle y^2=4n^2+4n+1-\frac{4(n+1)n}3 =4(n+1)n(1-\frac13)+1 =$
$\displaystyle\ =\frac{8(n+1)n}3+1 =\frac{16i(j-i)j(j+i)}3+1$. Thus

$\displaystyle y^2-1=(y-1)(y+1)=\frac{16i(j-i)j(j+i)}3$.

Using that $y$ is odd and letting $y=2t+1$, we obtain
$\displaystyle 2t(2t+2)=4t(t+1)=\frac{16i(j-i)j(j+i)}3$, hence

$\displaystyle \frac{t(t+1)}2=\frac{2i(j-i)j(j+i)}3$.

Obviously the latter has no solutions with $j>i>0$ (but I do not see how to show this elementary, perhaps it is difficult, perhaps I am overlooking something) since if it did have solutions then it would produce a solution to the original problem. It is easily seen that both sides of the latter equation are integers, indeed $\displaystyle\frac{t(t+1)}2$ is a triangular number, while if $3\not|\ i$ then $3|(j-i)j(j+i)$.

I posted the above equation as a separate question.