8
$\begingroup$

Prove that the equation $3^k = m^2 + n^2 + 1$ has infinitely many solutions in positive integers.

I have found that this is true for the first $k$'s from 1 to 7 except 3 and 6.

I have tried algebraic manipulation and induction too and it doesn't seem to work. I believe induction won't work since there are exceptions.

If I am correct, the numbers $m^2$ and $n^2$ can only be of the form $3a+1$.

Do you have any ideas about how I should proceed with this? I would love a few hints. Thanks.

$\endgroup$
1
  • $\begingroup$ Hint: proof that there is infinitely Pythagorean Triples, $z^2=m^2+n^2$ and $z^2=3^k-1$. $\endgroup$
    – AsdrubalBeltran
    Mar 29, 2016 at 14:15

2 Answers 2

Reset to default
6
$\begingroup$

EDIT(ELABORATION)

Note that $$(a^2+b^2)(c^2+d^2)=(ad+bc)^2+(ac-bd)^2 \tag{1}$$ Thus a product of two numbers that are a sum of $2$ squares is als0 a sum of two squares.

CLAIM

For all $t \in \mathbb{N}$, we have that $3^{2^{t}}-1$ is a sum of two squares.

PROOF

It is true when $t=1$ since $$3^{2}-1=2^2+2^2$$ Assume it is true when $t=a$. Note that for $t=a+1$, $$3^{2^{a+1}}-1=\left(3^{2^a} -1 \right) \left(3^{2^a}+1 \right)$$ By the inductive hypothesis, $3^{2^a} -1 $ is a sum of two squares. Also, $3^{2^a}+1$ is a sum of two squares from . Thus, the inductive hypothesis is true when $t=a+1$ from $(1)$. We are done. The result follows, as $k=2^{t}$.

$\endgroup$
10
  • $\begingroup$ $$3^{2^{n+1}} = 3^{2^n} \cdot 3^2$$ So? $\endgroup$
    – TheRandomGuy
    Mar 29, 2016 at 14:06
  • $\begingroup$ No, $3^{2^{n+1}}=\left(3^{2^n}\right)^2$ @Dhruv $\endgroup$
    – Thomas Andrews
    Mar 29, 2016 at 14:07
  • $\begingroup$ @ThomasAndrews That's equal to $$3^{2^n} \cdot 3^{2^n}$$ None of these is a sum of two perfect squares. I'm confused. $\endgroup$
    – TheRandomGuy
    Mar 29, 2016 at 14:11
  • $\begingroup$ But you won't want $3^{2^{n+1}}$ to be the sum of two squares, you want $3^{2^{n+1}}-1$ to be the sum of two perfect squares, given that $3^{2^n}-1$ is. @Dhruv $\endgroup$
    – Thomas Andrews
    Mar 29, 2016 at 14:13
  • $\begingroup$ Hint for the hint: what is $\Big(3^{2^{n+1}}\Big)^2 - 1$? $\endgroup$
    – shardulc says Reinstate Monica
    Mar 29, 2016 at 14:13
2
$\begingroup$

Hint: Prove that if $a^2-1$ is a sum of two square, then $a^{4}-1$ is a sum of two squares.

Thus if $3^{2k}-1$ is the sum of two squares for some $k$, then so is $3^{2^nk}-1$ for any $n$. (You can also prove if $3^{2k}-1$ is the sum of two squares, then so is $3^{k}-1$.)

The hint given by MXYMXY is just the case $k=1$.

The case $k=5$ also has this property, because $$3^{10}-1=(3^5-1)(3^5+1)=8 \cdot 11^2\cdot 61=(11\cdot 22)^2 + 22^2$$ is the sum of two squares. So $3^{5\cdot 2^n}-1$ is always the sum of two squares.

$\endgroup$
2
  • $\begingroup$ We also have $3^{26} = 253118^2 + 1574102^2 + 1$ and $3^{26} = 455002^2 + 1528018^2 + 1$. Solutions exist for many odd exponents of 3 as well, like 9, 13, 15, 19, 21, and 27. $\endgroup$
    – shardulc says Reinstate Monica
    Mar 29, 2016 at 14:40
  • $\begingroup$ They have to exist if they exist for even. For example, $3^{k}-1$ follows from $3^{2k}-1$, but not visa-versa when $k$ is odd. You need the even case to get the induction for an infinite class, but there are cases when $k$ is odd. $\endgroup$
    – Thomas Andrews
    Mar 29, 2016 at 14:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.