A Nice Diophantine Equation | 2nd method

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[TheEternalVortex]

Alternate solution that doesn't require checking any cases. Take (b+k)^2 - b^2 as in the video, then we get 2bk + k^2 = 12

This implies k is even. But if k >= 4 then the LHS is divisible by 8 but the RHS is not, so we must have k = 2. Then immediately we get b = 4.

[stardreamix]

Nice equation - good to know if we don't know the difference of 2 squares!

[goldfing]

a^2 - b^2 = 12

Since we have a Diophantine equation, a and b are both integers.

1.) a^2 and b^2 are non-negative square numbers.
2.) a^2 and b^2 are both even or both odd.
3.) a and b are both even or both odd.
4.) a^2 > b^2.
5.) abs(a) > abs(b).

There are not that many possibilities.
What two square numbers have a difference of twelve?
To my knowledge, only 4 and 16.

Any other two square numbers produce a different difference,
mostly larger than 12:

E.g. if both square numbers are even:
4 - 0 = 4, too little
16 - 4 = 12, ok
16 - 0 = 16, too much
36 - 16 = 20, too much

E g. if both square numbers are odd:
9 - 1 = 8, too little
25 - 9 = 16, too much
25 - 1 = 24, too much

Altogether, we get

a^2 = 16
a = +-4

and

b^2 = 4
b = +-2

[habeebalbarghothy]

Without k we can solve it !
(a-b)(a+b)=6x2

[ryanrahuelvalentine]

Bro make a video in perfect numbers. please.

[robertveith]

*@ SyberMath* -- 3rd Method: Let a = kb, for a rational number k. Substitute.
k^2b^2 - b^2 = 12
b^2(k^2 - 1) = 12
b^2 must equal 1 or 4 of the factors of 12.
If b^2 = 1, that leaves k^2 - 1 = 12, which has no rational solutions.
If b^2 = 4, then b = +/- 2. k^2 - 1 = 3, so k^2 = 4.
Then, k = +/- 2. And, a = +/- 4.
Solutions: (+/- 4, +/- 2)
*(Edit.)*