Your tricks are very nice,but you do not need any tricks here. I solved it in my head like this: multiply eq 1 with cd and eq 2 with bd and subtract them,you wil get: 2(cd)^2+(bd)^2+3(cd)-(bd)=0. Since (bd)^2>=bd for evry integer bd, cd must be 0(unless cd=-1 but then (bd)^2-bd=1 no sol). Now just solve for c=0 and d=0 it is easy.
@@yoav613 not really...squaring and factoring is as rudimentary as mult/sub. Also ...your bounding condition is more complex than discovering that d=0 from the author's way. Sure the author's remaining steps thereafter are more intricate, but alternatively he could have just plugged d=0 back into the original system. If one had also take the extra step to solve a with d...then the remaining is trivial assuming the exact same way you solved b,c. Though since you didn't solve a...your method might take 1-2 more steps to do it.
@@bait6652 ok i will not argue because anyone has his favorite methods to solve problems.at least for me it was very easy to solve this way and i solved it quickly in my head.
Note how the LHS of these equations similar to the real and imaginary parts of (a+id)(b+ic) But instead of using i, you use s = sqrt(-2). This gives us (a+sd)(b+sc) = 3 + s Taking norms on both sides gives that (a²+2d²)(b²+2c²) = 11 LHS is a product of integers, RHS is prime. So the only option is to have one of them be 1, the other 11. The norm being 1 means it's a unit, and in most rings of integers of imaginary quadratic number fields, that means the number itself is 1. We end up with the same solutions.
Great video, as always. Here is my favorite approach, without squaring (guess what, it leads to a sum of squares, kind of!): Multiply the first by cd, i.e. abcd - 2(cd)² = 3cd and the second one by bd, thus abcd + (bd)² = bd. Subtract the first from the second: (bd)² - bd + 2(cd)² + 3cd = 0 (just like yoav's approach) and complete the squares (multiply by 8 first): 8(bd)² - 8bd + 16(cd)² + 24cd = 0 or 2[4(bd)² + 4bd + 1] + (4cd)² + 2(4cd)3 + 3²= 2 + 3², hence 2(2bd-1)² + (4cd+3)² = 11, i.e. almost a sum of squares, with a small right hand side and easy to solve. Clearly, the only integer solutions are 2bd-1=±1, 4cd+3=±3, from which we get bd=0 or 1 and cd=0 (cd=-3/2 is rejected). For the case bd = 0 and cd = 0, combinations b = c = 0, b = d = 0 and c = d = 0 are all rejected (initial equations cannot hold). Thus, d = 0 and the initial equations are ab = 3 and ac = 1, giving the solutions (1,3,1,0), (-1,-3,-1,0). Similarly, the case bd = 1 and cd = 0 gives c = 0, hence ab = 3 (first equation), which give the solutions (3,1,0,1) and (-3,-1,0,-1).
I did this much the same way but I didn't factorise the polynomial that =11. It was still clear where d=0 that 1,3,1 (plus or minus) were the only solutions and when d=1 we have 3,1,0 (plus or minus).
Your tricks are very nice,but you do not need any tricks here. I solved it in my head like this: multiply eq 1 with cd and eq 2 with bd and subtract them,you wil get: 2(cd)^2+(bd)^2+3(cd)-(bd)=0. Since (bd)^2>=bd for evry integer bd, cd must be 0(unless cd=-1 but then (bd)^2-bd=1 no sol). Now just solve for c=0 and d=0 it is easy.
isn't it a similar style trick?
@@bait6652 no this is more standart way to solve this,and more easy way
@@yoav613 not really...squaring and factoring is as rudimentary as mult/sub. Also ...your bounding condition is more complex than discovering that d=0 from the author's way. Sure the author's remaining steps thereafter are more intricate, but alternatively he could have just plugged d=0 back into the original system. If one had also take the extra step to solve a with d...then the remaining is trivial assuming the exact same way you solved b,c. Though since you didn't solve a...your method might take 1-2 more steps to do it.
@@bait6652 ok i will not argue because anyone has his favorite methods to solve problems.at least for me it was very easy to solve this way and i solved it quickly in my head.
@@yoav613 it is impressive tho that you manage to mentally keep all the signs correctly to deduce cd
Nice one 👍
Please also do some questions on geometry calculus.
Its a request please.
Note how the LHS of these equations similar to the real and imaginary parts of (a+id)(b+ic)
But instead of using i, you use s = sqrt(-2).
This gives us (a+sd)(b+sc) = 3 + s
Taking norms on both sides gives that (a²+2d²)(b²+2c²) = 11
LHS is a product of integers, RHS is prime. So the only option is to have one of them be 1, the other 11.
The norm being 1 means it's a unit, and in most rings of integers of imaginary quadratic number fields, that means the number itself is 1.
We end up with the same solutions.
Wow nice one...is there a reason you write s instead of sqrt(2)i
@@bait6652 Just makes it easier to write over and over again.
I thought that squaring and subtraction would be the last way to try...
But it turned out to be a good solution...
Great video, as always.
Here is my favorite approach, without squaring (guess what, it leads to a sum of squares, kind of!):
Multiply the first by cd, i.e. abcd - 2(cd)² = 3cd and the second one by bd, thus abcd + (bd)² = bd.
Subtract the first from the second: (bd)² - bd + 2(cd)² + 3cd = 0 (just like yoav's approach) and complete the squares (multiply by 8 first):
8(bd)² - 8bd + 16(cd)² + 24cd = 0 or 2[4(bd)² + 4bd + 1] + (4cd)² + 2(4cd)3 + 3²= 2 + 3², hence 2(2bd-1)² + (4cd+3)² = 11, i.e. almost a sum of squares, with a small right hand side and easy to solve.
Clearly, the only integer solutions are 2bd-1=±1, 4cd+3=±3, from which we get bd=0 or 1 and cd=0 (cd=-3/2 is rejected).
For the case bd = 0 and cd = 0, combinations b = c = 0, b = d = 0 and c = d = 0 are all rejected (initial equations cannot hold). Thus, d = 0 and the initial equations are ab = 3 and ac = 1, giving the solutions (1,3,1,0), (-1,-3,-1,0).
Similarly, the case bd = 1 and cd = 0 gives c = 0, hence ab = 3 (first equation), which give the solutions (3,1,0,1) and (-3,-1,0,-1).
I did this much the same way but I didn't factorise the polynomial that =11. It was still clear where d=0 that 1,3,1 (plus or minus) were the only solutions and when d=1 we have 3,1,0 (plus or minus).