17
$\begingroup$

Are there solutions in integers $a,b>1$ to the following simultaneous congruences? $$ a^4\equiv 1 \pmod{b^2} \quad \mathrm{and} \quad b^4\equiv1 \pmod{a^2} $$

A brute-force search didn't turn up any small ones, but I also don't see how to rule them out.

$\endgroup$
5
  • $\begingroup$ This question actually results in the same congruences (just replace $a$ and $b$ by $m$ and $n$). One of the comments there suggests impossibility of that question's objectives; it might also show the impossibility of this one. $\endgroup$ Jul 18, 2013 at 2:17
  • $\begingroup$ @PeterKošinár your link directs to this post.. $\endgroup$ Jul 18, 2013 at 2:17
  • $\begingroup$ @Cameron Thanks! It should be correct now. $\endgroup$ Jul 18, 2013 at 2:19
  • $\begingroup$ @PeterKošinár yep! Works now. $\endgroup$ Jul 18, 2013 at 2:19
  • $\begingroup$ Related: Find $a,b$ for which $a^2\equiv-1\pmod b$ and $b^2\equiv-1\pmod a$ $\endgroup$ Apr 13, 2015 at 21:17

2 Answers 2

1
$\begingroup$

Maybe like this: First observe that b^2 | (a^4-1) and a^2 | (b^4-1). Hence, we can write a^4-1=kb^2 and b^4-1=ma^2, with k, m being positive integers.

So that gives then that a^4-1 = ((b^4-1)/m)^2-1=(b^8-2b^4+1)/m^2-1 = kb^2

That means that b^8-2b^4+1 = kb^2*m^2+m^2

Hence b^2(b^6-2b^2-km^2)=m^2-1

Now note that b^2 then should be a divider of m^2-1, cause b^6-2b^2-km^2 is still a integer. That means that m^2 = 1 mod b^2, with m and b being still integers. Hence, m^2 = 1 +lb^2, with l being integer. Substitution then gives that b^6-(2+lk)b^2-(k+l)=0

Now solve this equation for b^2 and observe that this gives two complex solutions and one real solution, the only thing to do is to solve that explicitly and try for integer values of l and k to get integer value of b, thereby m^2 also becomes integer and a divider.

$\endgroup$
0
$\begingroup$

I would think this could be attacked as follows. Write $b = b_1b_2b_3$ for integers $b_1,b_2,b_3$, such that $b_1^2 \mid (a-1)$ and $b_2^2 \mid (a+1)$ and $b_3^2 \mid (a^2+1)$. Hence the first condition is satisfied, i.e.,

$$ a^4-1 = (a-1)(a+1)(a^2+1) \equiv 0 \pmod{b^2}.$$

Write $a-1 = c_1b_1^2$ and $a+1=c_2b_2^2$ and $a^2+1=c_3b_3^2$ for integers $c_1, c_2,c_3$.

From this point, I would think there would be a number of interesting ways to solve or contradict the second condition

$$ b^4-1 = (b-1)(b+1)(b^2+1) \equiv 0 \pmod{a^2}.$$

Good luck!

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .