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1
vote
Recurrence relation with two variables
The solutions look like a mess, so it's not too surprising that you always end up with one. If we follow Iosif Pinelis in dividing all by the last constraint by $\lambda$ and substituting $r = \frac{1 …
9
votes
Solving functional equation $f(xy)=f(x+y)$ and Diophantine equations
$7 \sim 12$ via $3, 4$
$12 \sim 35$ via $5, 7$
$35 \sim 264$ via $11, 24$
$264 \sim 41$ via $8, 33$
$41 \sim 420$ via $20, 21$
$420 \sim 43$ via $15, 28$
$43 \sim 156$ via $4, 39$
$156 \sim 25$ via $ …