For any integer n, define F(n) = p ? ?

For any integer n, define F(n) = {p ? ? | n^4 + 1 = p^4 + 1} Then let P be the partition of the integers that contains all the sets F(n). That is: P = {F(n) | n ? ?} Then the cardinality of P (i.e. the number of elements in P) is

Transcript

00:01 Here we have f of n is defined as the set of all integers p such that n to the power 4 plus 1 equals p to the power 4 plus 1.

00:16 Now n to the power 4 plus 1 equals p to the power 4 plus 1 implies that p to the power 4 minus n to the power 4 equals 0.

00:27 And we can factorize this as p squared minus n squared times p.

00:33 Squared plus n squared equals 0 then we can factorize b squared minus n squared similarly as p minus n times p plus n times p plus n squared equals 0 so we must have that at least one of these factors must be 0 now if p minus n equals 0 we have p equals n if p plus n equals 0 we have p equals minus n and and and we have p equals minus n and and if p squared plus n squared equals 0, we have p equals n equals 0.

01:14 So then f of n can be written as f of 0 is the singleton set containing 0, and for n not equal to 0, f of n is the set containing minus n and n.

01:37 Next we'll find the cardinality of the set p, which consists of del, del, and n.

01:44 Limits f of n which are sets.

01:47 So this is a set of sets.

01:49 So p equals f of n for each integer is n in z.

01:56 Note that for n not equal to zero, f of n is the set minus n comma n, which is the same as f of minus n.

02:06 Because when we are dealing with f of minus n, we have minus n and minus of minus n, which is n.

02:12 So f of n and f of minus n are the same sets.

02:18 And since p is a set, it contains only unique elements...

Question

For any integer n, define F(n) = {p ? ? | n^4 + 1 = p^4 + 1} Then let P be the partition of the integers that contains all the sets F(n). That is: P = {F(n) | n ? ?} Then the cardinality of P (i.e. the number of elements in P) is

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