Is it true: $3k^2+1$ is a perfect square if and only if $k=1$ or $4$

Question

I'm checking the following conjecture: $3k^2+1$ is a perfect square if and only if $k=1$ or $4$. If it is not true counter example would be appreciated. Thanks in advance.

Answer 1

There are in fact infinitely many solutions:

$k_1=1, k_2=4, k_3=15, k_n=4k_{n-1}-k_{n-2}$

$3\cdot15^2+1=676=26^2$, etc.

Answer 2

This is not true. $3\times15^2+1=676=26^2$.

Answer 3

You can do this by solving the pells equation.

Let $3{k^2}+1={n^2}$ for some natural number n.Then the equation becomes a

pells equation$\implies$ ${n^2}-3{m^2}=1$.Then factorize L.H.S.

We get $(n+{\sqrt3}m)(n-{\sqrt3}m)=1$.Here m and n are solutions of the equation.

Now square both sides of the equation,you will get$\implies$

$({n^2}+3{m^2}+2{\sqrt3}mn)({n^2}+3{m^2}-2{\sqrt3}mn)=1$

Now see that the above equation is of the form ${\implies}$

$(I_1+{\sqrt3}I_2)(I_1-{\sqrt3}I_2)=1$

So $I_1\;\;and\;\;I_2$ are also solutions of that above equation.So if

you proceed like this, you will get the solutions for K.

By observing some solutions for K,we get the recurrence

$K_n=4K_{n-1}-K_{n-2}$.

Hence by solving the recurrence by characteristic equation for solving

recurrence we get ${\implies}$

$$K_n=\frac{1}{2{\sqrt3}}(({2+{\sqrt3}})^n-({2-{\sqrt3}})^n)$$