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 $11^2$).

Answer

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.

Attribution
Source : Link , Question Author : Angelo Mark , Answer Author : Will Jagy

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