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

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SUMMARY

The equation \((3a+b)^2 + 6a - 2b = 1544\) is analyzed to find the values of \(a\) and \(b\) in the natural numbers. By rearranging and simplifying the equation, the solution reveals that \(a + b = 38\). This conclusion is reached through systematic substitution and solving techniques, confirming that both \(a\) and \(b\) are integers that satisfy the original equation.

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Albert1
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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)
 

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