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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 …
Peter Taylor's user avatar
  • 7,226
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 $ …
Peter Taylor's user avatar
  • 7,226