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

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To solve the equation (3a+b)² + 6a - 2b = 1544 for natural numbers a and b, one approach is to rearrange and simplify the equation. By substituting values for a and solving for b, or vice versa, potential pairs can be identified. After testing various combinations, the values of a and b that satisfy the equation can be determined. Ultimately, the goal is to find the sum a + b. The solution confirms that a and b are natural numbers, leading to a specific value for a + b.
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$a,b\in N$

given:

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

find:

$a+b=?$
 
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Albert said:
$a,b\in N$

given:

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

find:

$a+b=?$

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

Thread 'Area of Overlapping Squares'

Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?
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