Find all primes $p$ and $q$ such that $p^2-2q^2=1.$

Question

Find all primes $p$ and $q$ such that $p^2-2q^2=1.$

My idea so far was to observe that since $2q^2$ is even, then $q^2$ must be odd or even. If $q^2$ is even, then $q$ is even and the only even prime is $2.$ Thus one pair of primes $(p,q)=(3,2).$

But, if $q^2$ is odd, then $q$ is also odd and $q=2d+1$ for some integer $d.$ Then,

$p^2-2q^2=1$ turns into $p^2=4(2d^2+2d)+3.$

From here I do not know how to proceed.

Any ideas or suggestions would be greatly appreciated.

Answer 1

From your last expression we have $$\tag1(p+1)(p-1)=2\cdot[\;2(2d^2+2d)+1\;].$$
If $p=2$ then you get from the original expression that $2q^2=3,$ which is impossible. Therefore $p>2$ and hence $p$ is odd. Then $p-1$ and $p+1$ are both even, which implies that their product $(p+1)(p-1)$ is divisible by $4$ which is clearly false by $(1).$ Therefore $q$ must be even and thus it must be equal $2,$ which implies that $p=3.$

Another approach: Since $p^2-2q^2=1$ then $(p+1)(p-1)=2q^2$ and the Unique Factorization Theorem implies that either $p+1=q^2\;\wedge\;p-1=2$ or $p+1=2q\;\wedge\;p-1=q$ but in either case $p=3$ and $q=2.$

Answer 2

Since $p^2-2q^2=1$, $p$ must be odd. Furthermore, $q$ must be even; if $q$ were odd, then $p^2-2q^2$ would be $3$ mod $4$. Thus, $q=2$ and then $p=3$.

Answer 3

HINT ON THE SEARCH OF ALL SOLUTIONS

As an elementary consequence of Dirichlet's theorem of units on number fields, we have for the quadratic cases the general solution of the equation of Pell-Fermat $x^2-dy^2=\pm1$ which concerns the units (a.e. elements of norm equal to $\pm1$) of the field $\mathbb Q(\sqrt d)$.

The field $\mathbb Q(\sqrt2)$ has as fundamental unit of its ring of integers $\mathbb Z[\sqrt 2]$ the number $1+\sqrt 2$ so it is known that all the integer solutions of the equation $$x^2-2y^2=1$$ are given by $$a_{2n}+b_{2n}\sqrt 2=(1+\sqrt 2)^{2n}$$ Here we are interested in all the $(a_{2n},b_{2n})$ being both primes.

Answer 4

$p^2-2q^2\equiv p^2+q^2\equiv 1\pmod{3}$.

But $a^2\equiv \{0,1\}\pmod{3}$ for all $a\in\Bbb Z$, so exactly one of $p,q$ is equal to $3$.

$q=3$ gives $p^2=19$, impossible. $p=3$ gives $(p,q)=(3,2)$.

Answer 5

You can easily checks for solutions in {2,3,5}.

For prime p $\geq$ 6, $p \equiv \pm1 \pmod6$ $ \Rightarrow p^2 \equiv 1 \pmod6$ .Therefore, $ p^2 - 2q^2 \equiv 1 -2 \equiv -1 \not\equiv 1\pmod6$

Hence, no solutions other than (3,2) exists.

Answer 6

One more approach (just for exercising).

We use the knowledge that $1=3^2-2 \cdot 2^2$ and rewrite the initial formula as $$ p^2 - 2q^2 = 3^2-2 \cdot 2^2 \tag 1$$ and then $$ p^2 - 3^2 = 2(q^2-2^2) \tag 2$$ Factoring gives $$(p-3)(p+3) = 2(q-2)(q+2) \tag 3$$

Now the lhs must be even (because the rhs is), thus $p$ must be odd (additionally: an odd prime by problem statement), and the lhs is divisible by at least $2^3=8$.

So the parentheses in the rhs must be even, and thus $q$ must be even and thus $q=2$ as the only even prime.

Then the rhs is zero, thus the lhs must be zero, thus $p=3$ and $(p,q)=(3,2)$ is the only solution.