Finding $a+b$ given $(3a+b)^2+6a-2b=1544$

[Albert1]

$a,b\in N$

given:

$(3a+b)^2+6a -2b =1544$

find:

$a+b=?$

 

[kaliprasad]

$a,b\in N$

given:

$(3a+b)^2+6a -2b =1544$

find:

$a+b=?$

Spoiler

$(3a+b)^2 + 2(3a-b) = 1544$
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

 

[Albert1]

Spoiler

$(3a+b)^2 + 2(3a-b) = 1544$
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

nice solution (Yes)