Determine all integers n≥2, satisfying n=a2+b2,where a is the smallest divisor of n different from 1 and b is an arbitrary divisor of n. Proposed by Walther Janous
Problem
Source: Austrian Mathematics Olympiad Regional Competition (Qualifying Round) 2017, Problem 4
Tags: number theory, Sum of Squares, Divisors, Divisibility
12.06.2018 05:39
Can the arbitrary divisor can be chosen as the smallest divisor?
12.06.2018 05:41
^ Why not?
12.06.2018 06:03
Notice both sides of the equation are divisible by b. This means b divides a2. But a is simply the smallest prime divisor of n, so b=1 or b=a or b=a2. b=1 fails. Taking b=a, we have n=2a2, so 2 divides n, and as 2 is the smallest possible value that a divisor can have, we conclude a=2, giving us n=8=22+22. Taking b=a2, we have n=a2(1+a2). But one notices that since a is prime, if a>2, then (1+a2) must be even, meaning n is even, meaning it is divisible by 2, contradicting a's minimality. Thus a=2. This gives us one more solution, n=20=22+42. So, n=8 and n=20 are the only solutions.
12.06.2018 06:11
@above Please explain this! Taking b=a2, we have n=a2(1+a2). But one notices that since a is prime, if a>2, then (1+a2) must be even, meaning n is even, meaning it is divisible by 2, contradicting a's minimality.
12.06.2018 06:15
IMO2019 wrote: @above Please explain this! Taking b=a2, we have n=a2(1+a2). But one notices that since a is prime, if a>2, then (1+a2) must be even, meaning n is even, meaning it is divisible by 2, contradicting a's minimality. "Contradicting a's minimality"="contradicting the fact that a is the smallest prime that divides n". We know a is the smallest prime that divides n, yes? But if a>2 then a2+1 would be even, because a is odd. So n is divisible by 2. But 2 is smaller than a! So this contradicts the definition of a, because nothing can be smaller! So the case fails!
12.06.2018 06:18
Oh thanks! Nice solution. I am so stupid.
12.06.2018 10:47
See my solution from here
12.06.2018 10:49
Oh, thanks for the link! I am locking this topic.