Let $a,b$ be primes. Find all primes such that $a^2+b^2=9ab-13$.
Whatever I've done is from parity checking. But I can't proceed with the case when both $a,b$ are odd primes. I tried with some modulo-chasing but couldn't complete.
2026-04-24 06:55:35.1777013735
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Find all primes such that $a^2+b^2=9ab-13$.
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Not a complete answer, just a few restrictions...
Proposition 1. For any prime $p>3 \Rightarrow 3 \mid p^2-1$ (from LFT) and $8 \mid p^2-1$ (from $(2k+1)^2 \equiv 1 \pmod{8}$). As a result $24 \mid p^2-1$.
For $a>3,b>3 \Rightarrow 24 \mid a^2-1$, $24 \mid b^2-1$ and $$24 \mid a^2 + b^2 -2=9ab-15$$ or $$8 \mid 3ab-5 \tag{1}$$ But, any prime $p>3$ is of the $p=4k+1$ or $p=4k+3$ form. None of $a,b$ can be of the same form at the same time:
- $a=4k_a+1,b=4k_b+1 \Rightarrow 3ab-5=3(4k_a+1)(4k_b+1)-5=12Q-2$ is not divisible by 4 and thus not divisible by 8.
- $a=4k_a+3,b=4k_b+3 \Rightarrow 3ab-5=3(4k_a+3)(4k_b+3)-5=12Q-22$ is not divisible by 4 and thus not divisible by 8.
As a result, either $$a=4k_a+1,b=4k_b+3 \color{red}{\text{ or }} a=4k_a+3,b=4k_b+1 \tag{2}$$ Going further: $$3ab-5=3(4k_a+1)(4k_b+3)-5=3(16k_ak_b+12k_a+4k_b)+4$$ reveals that both $k_a,k_b$ can't be odd or even at the same time.
We can also assume $a\leq b$ and from $a^2+b^2-9ab=-13<0$, by checking $x^2-9x+1<0$ where $x=\frac{b}{a}$, we have $$a\leq b < 9a \tag{3}$$
unlikely that this can be finished... ALL integer solutions are in two families, related by the mapping $$ (x,y) \mapsto (9x-y, x) $$ beginning with two "seeds," $$ (1,2), \; \; (7,1) , \;\; ( 62,7), \;\; ( 551, 62), \;\; $$ $$ (2,1), \; \; (17,2) , \;\; ( 151,17), \;\; ( 1342, 151), \;\; (11927, 1342), \;\; $$ it suffices to take $x,y > 0,$ as we cannot have a solution with $xy \leq 0$
For each orbits, we have separate linear recurrences ( by Cayley-Hamilton), $$ x_{n+2} = 9 x_{n+1} - x_n, $$ $$ y_{n+2} = 9 y_{n+1} - y_n. $$ $$ 2p, \; 17p, \; 151p, \; 1342, \; 11927p, \; 106001, \; 942082, \; 8372737p, \;74412551, \; 661340222, \; 5877649447p, \; 52237504801, \; 464259893762, \; 4126101539057, $$ Here we see the consecutive primes $17, 151$ which is probably the largest answer. $$ 1, \; 7p, \; 62, \; 551, \; 4897, \; 43522, \; 386801, \; 3437687, \; 30552382, \; 271533751, \; 2413251377, \; 21447728642, \; 190616306401, \; 1694099028967, $$
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