# 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 $$11^2$$).

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

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.