<s>[INST] 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.
If we ignore one term, we can make the remaining terms squares.  For example, $3k+1$ and $4k+1$ are both squares if $k=56$ (they are $13^2$ and $15^2$); $3k+1$ and $6k+1$ are both squares if $k=8$ (they are $5^2$ and $7^2$); $4k+1$ and $6k+1$ are both squares if $k=20$ (they are $9^2$ and $11^2$).
 [/INST] 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)
</s>