I was wondering too, I'm not convinced. X and y don't need to be poitive integers so I don't understand how he came to this? That said I suppose the found values are correct, just maybe not all the possible correct answers ?
@@ProfessorBorax we have to prove that x and y are integers: √b=√603-√a b=a+603-2√(603a) Then 2√(603a) is an integer c Thus by squaring 4*603*a=c² Since 603=3²*67 and 4=2² and 2,3,67 are prime numbers So 2,3,67 divide c² and divide c too Hence their product divide c (prime numbers) c=2*3*67x and c²=4*9*67²x², x integer Then 4*603*a=4*9*67²x² 4*9*67*a=4*9*67²x² And finally a=67x with x integer We do the same with b.
@@itsmetrendy8471we have to prove that x and y are integers: √b=√603-√a b=a+603-2√(603a) Then 2√(603a) is an integer c Thus by squaring 4*603*a=c² Since 603=3²*67 and 4=2² and 2,3,67 are prime numbers So 2,3,67 divide c² and divide c too Hence their product divide c (prime numbers) c=2*3*67x and c²=4*9*67²x², x integer Then 4*603*a=4*9*67²x² 4*9*67*a=4*9*67²x² And finally a=67x with x integer We do the same with b.
Well, 0 is a non-negative integer. Note: Depending on the given condition, 0 might be considered as a natural number. So we can't entirely say 0 is not a natural number.
@@SpencersAcademy I looked it up àd it seems it's not 100% clear, but I also remember that N is 1, 2, 3, and if we want to include 0 we add that in the notation "No". Still it's probably good to include the answer and maybe add a note: if 0 is included ... bla bla bla ^^
According to the 1st Peona axiomes, 0 is a natural number. Defined by: 1- S(0)=1 , successor of 0 is 1. 2- S(n)=0 is false, there is no natural number whose successor is 0.
Haven't watched yet. My guess would be to start from √(A+√B)= √(A+√(A^2-B))/2+√(A-√(A^2-B))/2. Since we need A^2-B to be perfect square this is guaranteed if A and √B are part of pythagoric triplets with A cathete and √B a whole number too Pyhtagoric triplets : m^2-n^2; 2mn; m^2+n^2 where m and n are relatively prime. Expressing 603 as prime factors 603=3*3*67 for this we need pairs of lengths of 1 cathete and hypothenuse adding up to 67, because we can scale then later to match 9*67 So m^2+n^2+2mn=9×67 Or m^2+n^2+m^2-n^2=9*67 So (m+n)^2=9*67 Or 2m^2=9*67 Both cases are impossible as 67 is prime
@@SpencersAcademy but failed to get the obvious result lol Should have considered √67 as a factor and then 3,4,5 can get 9=4+5 √(5+3)/2+√(5-3)/2=2+1 solution would be achievable. This approach would also ignore the 0 ,3 solution as triangles don't have 0 size edges (unless they are degerate case n=0)
