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Starting from a math problem involving a square and a unit circle, I used some elementary algebraic transformations and discovered that for any arbitrary $u, v$, if we define $a, b, c$ as follows:

$$a = (u^{16} - 20u^{12}v^4 - 26u^8v^8 - 20u^4v^{12} + v^{16})^2$$ $$b = 64u^4v^4(u^4 + v^4)^2(u^4 - v^4)^4$$ $$c = (u^8 + 6u^4v^4 + v^8)^4$$

Then we always have $a + b = c$.

Question 1: With this parametric form, is it possible to generate infinitely many mutually coprime triples $(a, b, c)$ that satisfy $c > \text{rad}(abc)$?

Update: Following some simple algebraic manipulations (substituting $u^2 \to u$ and $v^2 \to v$ as suggested by Guruprasad) and then setting $u, v$ based on a Pythagorean triple, the identity $a + b = c$ can be re-parametrized using the generators $p$ and $q$ as follows:

$$a = \left[ (p^2-q^2)^8 - 20(p^2-q^2)^6(2pq)^2 - 26(p^2-q^2)^4(2pq)^4 - 20(p^2-q^2)^2(2pq)^6 + (2pq)^8 \right]^2$$

$$b = 64 \cdot (p^2-q^2)^2 \cdot (2pq)^2 \cdot (p^2+q^2)^4 \cdot \left( (p^2-q^2)^2 - (2pq)^2 \right)^4$$

$$c = \left[ (p^2-q^2)^4 + 6(p^2-q^2)^2(2pq)^2 + (2pq)^4 \right]^4$$

Based on a SageMath implementation with the conditions $(p,q) \in \mathbb{Z}_{>0}^2, q < p \le N, \gcd(p,q)=1, p \not\equiv q \pmod{2}$, it appears that for every such $p$, there exists at least one set of $(a, b, c)$ that satisfies the abc-hit condition ($q > 1$)."

Question 2: Infinite Recursive "Pythagoreanization" and Degree Reduction

If we repeatedly apply the degree reduction $p^2 \to p, q^2 \to q$ and then re-parametrize $p$ and $q$ themselves as a new Pythagorean triple ($p = r^2 - s^2, q = 2rs$) infinitely many times, what would be the result?

Empirical evidence from further iterations—strictly maintaining the conditions $\gcd(p,q)=1$ and $p \not\equiv q \pmod{2}$ to ensure primitivity at each step—suggests that the ratio $\frac{\ln(b)}{\ln(\text{rad}(b))}$ asymptotically approaches 4. Magnitude-wise, this implies that the recurrence generates infinitely many identities behaving like:

$$X^2 + \text{rad}(b)^4 = Z^4$$

Is there a rigorous mathematical proof or a known theorem that formally explains this limit, namely $\lim_{k \to \infty} \frac{\ln(b_k)}{\ln(\text{rad}(b_k))} = 4$?

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    $\begingroup$ This parametrization can be simplied further by taking $m=u^2$ and $n=v^2$ $\endgroup$ Commented Mar 27 at 11:06
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    $\begingroup$ @Guruprasad or even $u^4,v^4$. $\endgroup$ Commented Mar 27 at 14:32
  • $\begingroup$ @Wojowu I just saw that too. $\endgroup$ Commented Mar 27 at 14:42
  • $\begingroup$ @Wojowu , why I don't take $u^4$, $v^4$ because parametrization in a OP's post satisfies $p^2+q^2=r^4$. If we take $u^4$ ,$v^4$ then it doesn't satisfy $p^2+q^2=r^4$. $\endgroup$ Commented Mar 27 at 14:47

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Yes, there are infinitely many good abc triples coming from your identity.

From your first identity working as homogeneous polynomials we have $\deg({c})=32$ and $\deg{(\operatorname{rad}(abc))=34}$. This gives the "polynomial" quality $\frac{32}{34}<1$.

$b$ is divisible by $uv(u^4-v^4)$.

Set $v=1$ and $u=B^k$ for positive integers $B,k$.

The radical of $b$ is $ < 2\cdot B \cdot (B^{4k}-1) (B^{4k}+1)$

So far $u,v$ the decrease degree in the radical of $b$ by 2 giving near $1$ quality.

For arbitrary large integers $m$ and fixed $B$, we have $m^2$ divides $(B^{4k}-1)$ for some $k$ and this reduces the radical by a factor of at least unbounded $m$ which gives quality greater than one.

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    $\begingroup$ More explicitly, for your choice of $u,v$, we get $c\approx B^{32k}$ and $rad(abc)\lesssim rad(B^{4k}-1)B^{28k+1}$. Taking e.g. $B=2$ and $3\mid k$ we get $rad(B^{4k}-1)<B^{4k}/3$ so the inequality for large $k$. $\endgroup$ Commented Mar 28 at 15:03
  • $\begingroup$ @Wojowu Empirical evidence from further iterations suggests that the ratio $\frac{\ln(b)}{\ln(\text{rad}(b))}$ asymptotically approaches 4. Magnitude-wise, this implies that the recurrence generates infinitely many identities behaving like $X^2 + \text{rad}(b)^4 = Z^4$. Is there a rigorous mathematical proof or a known theorem that formally explains this limit? $\endgroup$ Commented Mar 28 at 17:18
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    $\begingroup$ @ĐàoThanhOai Look at the elliptic curve over the rationals: $X^4+k b^4=Y^2$ $\endgroup$ Commented Mar 28 at 17:47
  • $\begingroup$ @joro If we fixed $q=const$ or $p=const$, we have $deg(c)=32$ and $deg(rad(abc))=33$ $\endgroup$ Commented 2 days ago
  • $\begingroup$ @ĐàoThanhOai I am not working with p,q. If you fix v=1 u=B^k, then deg(rad(abc))=33 indeed, but the radical is divisible by B^k and rad(B^k)<=B and we have degrees of freedom to change B,k and get the radical divisible by m^r for integers m,r. $\endgroup$ Commented yesterday

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