2014/024) Given a and b natural numbers

Given a and b natural numbers

(3a + b)^2 + 6a – 2b = 1544

find a +b



Solution

(3a + b)^2 +  6a – 4b = 1544

add 4b+1 to both sides to get

(3a + b)^2 + 2(3a + b) + 1 = 1544 + 4b

or (3a + b + 1)^2 = 1545 + 4b

as 1545 mod 4 = 1 solution may exist


so we need to take odd squares above 1545

(3a+b+1)=41=>1545+4b=1681=>b= 34

this gives a = 2 or a + b = 36

(3a+b+1)=43=>1545+4b=1849=>b=76 ( it should be be < 3a + b + 1 for being positive)

so a+b = 36 is the only solution