A Nice Diophantine Equation | 1st method

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[GirishManjunathMusic]

a² – b² = 12 → |a| > |b|
(a + b)(a – b) = 12

Factors of 12: (±12, ±1), (±6, ±2), (±4, ±3)
of these, the only two pairs that work as (a + b) (a – b)
are ±6 and ±2.
Thus, a = ±4, b = ±2.

Cross-checking:
a² = 16; b² = 4; 16 – 4 = 12

[TheEternalVortex]

Since a and b are squared it's obvious that the negative of either or both of them will give a solution, so you can just focus on the case where both are positive, which means a + b > a - b so this gives 4 times fewer cases to check

[dwm]

Try it with 14, and in each case one factor is odd, the other even (1, 14) or (2, 7). So no whole number solution.
The constant on the RHS needs to be a multiple of 4.

OR The RHS can be odd

If (a + b).(a - b) = 35, for example
we get factor pairs (1, 35) which leads to a = 18, b = -17, etc
and (5, 7) which leads to a = 6, b = 1, and negatives.

[ejdansu]

This makes a lot of sense.

Thank you so much.

[wes]

(a–b)(a+b)=2*6. What if a=4 and b=2 or a=–4 and b=2 or b=–2 and a=4?

[vladimirkaplun]

a +b and a-b have same evenness hence 1, 12 and 3, 4 do not do at a glance

[chaosredefined]

Different approach, because you only really had the one.

a^2 - b^2 = 12
a^2 = b^2 + 12

WLOG assume a and b are both positive. (If (a, b) is a solution, then (-a, b), (a, -b) and (-a, -b) are all solutions as well). We automatically know that a > b. So, a - b > 0. Let's define c = a - b. So, a = b + c, and c > 0. Let's swap that in:
(b + c)^2 - b^2 = 12
b^2 + 2bc + c^2 - b^2 = 12
2bc + c^2 = 12
c(2b + c) = 12

This means that (c, 2b+c) = (1, 12), (2, 6), (3, 4). Note that we can't have the reverse of any of those since since 2b + c > c.

Furthermore, note that c can't be odd. If it is, then 2b+c is also odd, and so we have odd x odd = even. Therefore, the only possibility is (2, 6).

So, c = 2, 2b+c = 6. And 2b = 6-c = 6-2 = 4, so b=2. Finally, c was defined as a - b, so a - b = 2. Since b = 2, this means that a = 4.

Final solutions: (4, 2), (-4, 2), (4, -2), (-4, -2)

[pawelpap]

Please stop posting incorrectly stated problems. Your first slide should contain all the information. You cannot pull additional constraints as you go.