Solving A Nice Diophantine Equation

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❤️ a + 2ab + b = 52, find a + b

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[timothybohdan]

I got 16 integer solutions for (a, b) = {(-53, -1) (-18, -2) (-11, -3) (-8, -4) (-4, -8) (-3, -11) (-2, -18) (-1, -53) (0, 52) (1, 17) (2, 10) (3, 7) (7, 3) (10, 2) (17, 1) (52, 0). This gives 8 solutions for a+b = {-54, -20, -14, -12, 10, 12, 18, 52}.

[nasrullahhusnan]

a+2ab+b=52
a(1+2b)+½(1+2b)-½=52
(2b+1)(a+½)-½=52
½(2a+1)(2b+1)-½=52
(2a+1)(2b+1)=105
Note that the equation is cyclical and 105=1×3×5×7. W/o loss of generality we assume a<b. Hence
• 2a+1=1 --> a=0, b=52
2a+1=-1 a=-1, b=-53
• 2a+1=3 --> a=1, b=17
2a+1=-3 --> a=-2, b=-50
• 2a+1=5 --> a=2, b=10
2a+1=-5 --> a=-3, b=-11
• 2a+1=7 --> a=3, b=7
2a+1=-7 ---> a=-4, b=-8
Therefore
(a+b)={10, 12, 18, 52-53, -50, -11, -8}

[Qermaq]

I can find 8 unordered pairs all with unique sums. We see how you do it....

Ok, so I got 0 + 52 = 52, 1 + 17 = 18, 2 + 10 = 12, 3 + 7 = 10, -1 + -53 = -54, -2 + -18 = -20, -3 + -11 = -14 and -4 + -8 = -12. (Each of these is either a + b = ? or b + a = ?.) But I took a much less efficient route than you.

[swa]

What a nonsense, there are infinitely many solutions