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As an example:

$429^2+101^2=394\cdot 493$. 394 is the reverse of 493.

Are there infinitely many

$a^2+b^2$ which are the product of a number and his reverse?

a,b positive integers

Second question:

Are there infinitely many

$a^2+b^2$ which are the product of two composite numbers one the reverse of the other?

Obviously I asked this because it is connected with ec primes

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    $\begingroup$ Let $p,q$ be primes having $1\bmod4$ congruence and $q$ be reverse of $p$ (mathworld.wolfram.com/Emirp.html). The product $pq$ has $a^2+b^2$ representation. $\endgroup$
    – Turbo
    Commented Feb 3, 2023 at 19:39
  • $\begingroup$ @LSpice what if p and q are not primes? $\endgroup$
    – Enzo Creti
    Commented Feb 3, 2023 at 19:43
  • $\begingroup$ Re, then it depends quite a bit on the factorisations of $p$ and $q$. (I deleted a comment in response to @Turbo because I missed their requirement that both $p$ and $q$ be $1$ modulo $4$.) $\endgroup$
    – LSpice
    Commented Feb 3, 2023 at 19:44
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    $\begingroup$ This question would likely be a better fit for math.stackexchange.com; I wouldn't put it in the domain of 'research mathematics' per se (with a few rare exceptions, questions about digital representations and digit patterns tend not to be). $\endgroup$
    – Steven Stadnicki
    Commented Feb 3, 2023 at 20:02
  • $\begingroup$ One that doesn't fit the patterns mentioned so far: $15\times51=765=729+36=441+324$ ($=27^2+6^2=21^2+18^2$). $\endgroup$
    – Gerry Myerson
    Commented Feb 4, 2023 at 5:03

2 Answers 2

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Expanding on the comment by @Turbo:

If \begin{gather*} j=k^2+l^2 \\ s=m^2+n^2 \end{gather*} then by the Brahmagupta–Fibonacci identity we have $$ sj=(km+ln)^2+(km-ln)^2. $$ Therefore, if $s$ and $j$ are reverse pairs of this form, their product has the desired property.

It is very easy to construct infinitely many numbers $j$, $s$ of this form, for example there are infinitely many pairs which have only two non-zero digits, i.e. $s=u^2 \cdot 10^{2v}+w^2$ and $j=w^2 \cdot 10^{2v}+u^2$ for distinct non-zero digits $u$, $w$.

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If you don't require the two composite numbers to be distinct, try the square of any palindrome starting and ending in $5$.

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