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Find all positive integers $a$ and $b$ so that $a^4-21a^2b^2+4b^4$ is a divisor of $25137$.

I am stuck with this math problem. I know that: $$a^4-21a^2b^2+4b^4=(a^2+2b^2-5ab)(a^2+2b^2+5ab)$$ and $$25137=3^3\cdot7^2\cdot19$$ but then I don't know how to continue. Could you please help?

Bill Dubuque
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LE Thai-Hung
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    Welcome to Math SE. Note that if $d = \gcd(a^2+2b^2-5ab,a^2+2b^2+5ab)$, then $d$ divides the difference of the factors, i.e., $d\mid 10ab$. – John Omielan May 12 '24 at 02:56
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    The comment of @JohnOmielan represents an elegant refinement that I would not have thought of. My (inelegant) approach would be to let the factors be represented by $~F_1,F_2,~$ and attempt to apply the following analysis to each of $~F_1,F_2.~$ Use a $~\pmod{n}~$ argument, where $~n \in {3, 3^2, 3^3, 7, 7^2, 19},~$ to try to identify all ordered pairs $~(a,b) ~: ~a,b \in \Bbb{Z^+},~$ such that $~F_1~$ and/or $~F_2~$ is congruent to $~0, \pmod{n}.~$ – user2661923 May 12 '24 at 03:03
  • Re last comment, notice that you can easily find (for example) all quadratic residues $~\pmod{19},~$ by simply examining the set $~{0^2, 1^2, \cdots, 9^2}.~$ Note that (for example) $~(r + 19k)^2 \equiv r^2 \pmod{19},~$ and that $~(19-r)^2 \equiv (-r)^2 \equiv (r)^2 \pmod{19}.$ – user2661923 May 12 '24 at 03:07
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    one cannot use the 3 or the 7. These are not quadratic residues $\pmod{17}.$ If the polynomial is divisible by $3$ then both $a,b$ are divisible by 3, and the quartic is divisible by $81.$ Same comment for $7$ and $2401$ – Will Jagy May 12 '24 at 03:20
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    Meanwhile, now emphasizing $a,b > 0$ we can only allow $a^2 - 5ab + 2 b^2 = 1$ and $a^2 + 5ab + 2 b^2 = 19.$ However, this requires $10ab = 18$ which is impossible in integers, see the comment by @JohnOmielan . So that's it, no positive solutions, likely the only solution is $a = \pm 1, b=0$ – Will Jagy May 12 '24 at 03:26
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    @WillJagy FYI, note that $a=1$ and $b=2$ gives $a^2+2b^2-5ab=1+8-10=-1$ and $a^2+2b^2-5ab=1+8+10=19$, so their product of $-19$ divides $25137$ and, thus, $(1,2)$ is a positive integer solution. – John Omielan May 12 '24 at 03:53
  • Isn't this question now settled? By Will's reasoning, we must have $a^2+2b^2\pm5ab\in S:={\pm1,\pm19}$. By John's observation the difference of the two factors is $10ab$. The only occasion when the difference of two numbers from the set $S$ is a multiple of $10$ is (observing that $a^2+2b^2+5ab>3$): $a^2+2b^2-5ab=-1$, $a^2+2b^2+5ab=19$. Hence $10ab=20$, and... – Jyrki Lahtonen May 12 '24 at 06:55
  • $a^2+5ab+2b^2=(1/4)(4a^2+20ab+8b^2)=(1/4)(2a+5b)^2-17b^2)$ might be helpful. – Gerry Myerson May 12 '24 at 06:55
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    @JohnOmielan I see what you mean, I mentally discarded negative divisors unless they would both be negative. Can't win 'em all. – Will Jagy May 12 '24 at 13:45
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    @JyrkiLahtonen As neither $p=3$ or $p=7$ may be factors of $a$ or $b$, note that for both of these values of $p$ we have $a^4-21a^2b^2+4b^4\equiv (a^2)^2+(2b^2)^2\equiv 0\pmod{p}$ has no solutions as the quadratic residues of $p=3$ is just $1$, while for $p=7$ they are just $1$, $2$ and $4$, with neither one having a sum congruent to $0$. Although the general protocol for community wiki posts is that most members are allowed to edit & update them, I'm unsure if that's appropriate for your answer. As such, you may wish to update it yourself with this if you think it's useful. – John Omielan May 12 '24 at 14:06
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    You are welcome to edit the CW answer @JohnOmielan. And fix my English ;-) – Jyrki Lahtonen May 12 '24 at 17:26
  • @JyrkiLahtonen Thanks for your feedback. As you suggested, I've edited the CW answer to include the details of my comment to you. However, as for "fix my English", I didn't find any problems with it, apart from an almost trivial issue of where I thought there should be an additional comma, so I didn't make any other changes. – John Omielan May 12 '24 at 22:48
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    @JohnOmielan about possible English errors, Jyrki got his Ph.D. at Notre Dame in Indiana – Will Jagy May 12 '24 at 22:57

2 Answers2

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Compiling an answer from the comments under main.

  • As stated later by John Omielan, neither $p = 3$ or $p = 7$ may be factors of $a$ or $b$, as then $p^4$ would be a factor of the quartic, which is not possible. Note also that for both of these values of $p$, we have $a^4 - 21a^{2}b^{2} + 4b^4 \equiv (a^2)^2 + (2b^2)^2 \equiv 0 \pmod{p}$ has no solutions as the quadratic residues of $p = 3$ is just $1$, while for $p = 7$ they are just $1$, $2$ and $4$, with neither one having a sum congruent to $0$.
  • As explained by Will Jagy, either of the factors $a^2+2b^2\pm5ab$ is divisible by $3$ only when $3\mid a$ as well as $3\mid b$. But this would imply that the quartic is a multiple of $3^4$, which is a contradiction. If you are reluctant to do a brute force check, this also follows from the fact that $3$ is a quadratic non-residue modulo $17$ (Gerry Myerson's comment also explains where the modulus $17$ comes from).
  • The same applies to divisibility by $7$. Again, the quartic can be divisible by $7$ only when both $a$ and $b$ are, but then the quartic would be divisible by $7^4$.
  • So the factors $a^2+2b^2\pm5ab$ must both belong to the set $$S=\{\pm1,\pm19\}.$$
  • When constrained to have $a,b>0$, it follows that $a^2+2b^2+5ab\ge8$. As the difference of the two factors is $10ab$ (first observed by John Omielan), we are forced to conclude that $10ab=20$ (or $10ab=0$, again violating the positivity assumption).
  • It follows that the only solution is $a=1,b=2$, and this checks out (another comment by John Omielan).
John Omielan
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Jyrki Lahtonen
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Let $a$ and $b$ be positive integers such that $$a^4-21a^2b^2+4b^4\mid3^3\cdot7^2\cdot19.\tag{1}$$ Note that for $p\in\{3,7\}$ we have $p\equiv3\pmod{4}$ and so if $$a^4-21a^2b^2+4b^4\equiv(a^2)^2+(2b^2)^2\equiv0\pmod{p},$$ then $a^2\equiv2b^2\equiv0\pmod{p}$, and hence $a\equiv b\equiv0\pmod{p}$. Then also $$a^4-21a^2b^2+4b^4\equiv0\pmod{p^4},$$ contradicting the divisibility assumption in $(1)$. It follows that $$a^4-21a^2b^2+4b^4\mid19.$$ We can factor the left hand side as $$a^4-21a^2b^2+4b^4=(a^2+5ab+2b^2)(a^2-5ab+2b^2),$$ into a product of two factors that differ by $10ab$, a multiple of $10$. Then the product equals $\pm19$ and so $ab=2$, because $a$ and $b$ are positive. Then a quick check of the two remaining options shows that $(a,b)=(1,2)$ is the unique solution.

Servaes
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    This seems to be the same as the solution sketched in the comments yesterday (summarized in Jyrki's answer). Why repeat it (without attribution)? – Bill Dubuque May 12 '24 at 21:40