- Thread starter
- #1
Albert
Well-known member
- Jan 25, 2013
- 1,225
$a,b\in N$
given:
$(3a+b)^2+6a -2b =1544$
find:
$a+b=?$
given:
$(3a+b)^2+6a -2b =1544$
find:
$a+b=?$
$a,b\in N$
given:
$(3a+b)^2+6a -2b =1544$
find:
$a+b=?$
nice solution$(3a+b)^2 + 2(3a-b) = 1544$
add 4b+1 to both sides to get
$(3a+b)^2 + 2(3a+b) + 1 = 1545 + 4b$
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
$(3a + b + 1) = 43 => 1545 + 4b = 1849 => b= 76$ too large
so a+b = 36 is the only solution