Simultaneous perfect squares.

Рет қаралды 18,342

Күн бұрын

Пікірлер: 57

@mrphlip 3 жыл бұрын

I took a more number-theory approach: As in the video, looking at the cases where a,b nonzero. Say that a²+4b=m², and for simplicity say that m>=0. Since their difference is even, a and m must have the same parity. Suppose b>0... this means m>|a|, but since they have the same parity, this means m>=|a|+2. This puts a lower limit on b, as: b = (m²-a²)/4 >= ((|a|+2)²-a²)/4 = (a²+4|a|+4-a²)/4 = |a|+1 Alternatively, if b0, then |b| >= |a| + 1 If b= |a| - 1 (noting that the negative version is less strict, so we would expect our negative cases to have more options to work with) We can also, by symmetry, make the same argument with a and b reversed. And so, we have four main cases: Case 1: a and b are positive This means that a>=b+1 and b>=a+1, which is a contradiction, no solutions in this quadrant. Case 2: a positive, b negative This means a>=-b+1 and -b>=a-1, which squeezes our range down to a single value, so b=1-a. Plugging this in to our two formulae, we have: a²+4b = a² - 4a + 4 = (a - 2)² b²+4a = (1-a)² + 4a = a² - 2a + 1 + 4a = a² + 2a + 1 = (a + 1)² So both of these are perfect squares for any a, so we have another infinite family of solutions. Case 3: a negative, b positive By symmetry to case 2, the same line of solutions b=1-a continues through this quadrant. Case 4: a and b are negative This means that -a>=-b-1 and -b>=-a-1 which allows a range of 3 possible values: b=a-1, b=a, or b=a+1. Case 4a: b=a Plugging this into our formulae, we want a² + 4a to be a perfect square. However, we know that a² + 4a + 4 is a perfect square, as it's (a+2)², so we must have two perfect squares that differ by 4. The only two perfect squares that differ by 4 are 0 and 4 So we must have a² + 4a = 0 a = 0 or -4 We're only looking at the nonzero case, so we have a=b=-4, which is a valid solution. Case 4b: b=a+1 Plugging this into our formulae, we have: a² + 4b = a² + 4a + 4 = (a + 2)² b² + 4a = (a+1)² + 4a = a² + 6a + 1 The first is always a perfect square, so just need to look at the second. We know that a² + 6a + 9 is a perfect square, as it's (a+3)², so we must have two perfect squares that differ by 8. The only perfect squares that differ by 8 are 1 and 9 So we must have a² + 6a + 1 = 1 a = 0 or -6 We're only looking at the nonzero case, so we have a=-6, b=-5, which is a valid solution. Case 4c: b=a-1 By symmetry with case 4b, we have a=-5, b=-6. So, to sum up, we have 3 infinite families and 3 point solutions: a,b = (0,n²), (n²,0), (n,1-n), (-4,-4), (-5,-6), (-6,-5) for all integers n

@mrphlip 3 жыл бұрын

And now, hopefully that I've typed that whole thing out at 3am, my brain will let me actually sleep...

@megauser8512 3 жыл бұрын

@@mrphlip Nice!

@hanibahout 3 жыл бұрын

Super

@prag9582 5 ай бұрын

Amazing!

@goodplacetostop2973 3 жыл бұрын

17:59 Wishing you the ability to understand fast and easy, and the power to excel in your studies. Have a great week at school!

@rtistkalpanayadav7929 3 жыл бұрын

2nd comment

@MrRyanroberson1 3 жыл бұрын

4:00: a=b=-4 is a solution, this produces the desired zero in the radical

@Mephisto707 3 жыл бұрын

3:50 I don't think your setup (a and b not zero) precludes the discriminant being equal to zero (e. g., if b = - a^2/4)

@Falanwe 3 жыл бұрын

It doesn't, but the roots being different is not used anywhere either. Note the solution (-4, -4) actually corresponds to a zero discriminant ^^

@GreenMeansGOF 3 жыл бұрын

That’s what I was about to comment.

@n8cantor 3 жыл бұрын

Once we take care of the case when one root is equal to +/-1, you can simplify the alpha = 2 and alpha = -2 cases by recognizing that beta must also equal either 2 or -2 by the 1

@bosorot 3 жыл бұрын

I did this different and easy way. Assume (a+c)^2 = a^2+2ac+ c^2 , normal binomial for perfect square . a^2+4b is a perfect square . then 4b=2ac+c^2 (first equation) . and do the same method for b^2+4a . get pretty much like alike 4a=2bc+c^2 (second equation). Solve first and second equation for c = -2 . put c back to (either 1 or 2 equation )get a=1-b or b=1-a . Then I get indefinite set of integer (a, b) to satisfy the condition .

@jimschneider799 10 ай бұрын

2 years late but, @3:58 - I have a counterexample: a = b = -4: a^2 + 4*b = 0, and b^2 + 4*a = 0, both of which are perfect squares. This gives both quadratics a single (double) root at x = 2.

@matheusmeireles1781 3 жыл бұрын

13:22 Why (2a + 5)^2 is not possible to be 25? 2a + 5 can be 5 or -5, 5 is not possible cause we ser up that a is different than 0, but if 2a + 5 is -5, a can be -5 therefore b is -6, what is another solution (-5 , -6).

@michielhorikx9863 3 жыл бұрын

Presumably because this only works under the assumption that |a| >= |b|, which is false in that case. He'd be looking for the other solution (-6, -5).

@matheusmeireles1781 3 жыл бұрын

@@michielhorikx9863 Oh that´s true, I forgor

@gniedu 3 жыл бұрын

alpha = 1 yields 1+beta = - a beta = -b This leads to b = 1 + a So a² + 4b becomes a² + 4a + 4 = (a+2)², always a perfect square. And b² + 4a becomes a² + 6a + 1 = (a+3)² - 8. The only two perfect squares that differ by 8 are 1 and 9. So (a+3)² = 9, which makes a+3 = +- 3. a + 3 = 3 leads to a=0, already covered. a + 3 = -3 leads to : a = -6 and b = -5, another solution. Also the symmettic pair a = -5 and b = -6 _________________________________________________________________________________________________________________________________________ alpha = -2 yields: -2 + beta = -a -2beta = -b This leads to b = 4-2a So a² + 4b becomes a² - 8a + 16 = (a-4)², always a perfect square And b² + 4a becomes 4a² - 12a + 16 = (2a-3)²+ 7. The only two perfect squares that differ by 7 are 9 and 16. So (2a-3)² = 9 which makes 2a-3 = +-3 2a - 3 = -3 makes a =0, already covered 2a - 3 = 3 leads to: a = 3 and b = -2, another solution. Also the symmetric pair a = -2 and b = 3

@Minskeeeee 3 жыл бұрын

alpha = 1 leads to (-5,-6) and (-6,-5) alpha = -2 leads to no solution

@Macieks300 3 жыл бұрын

Wow, this is an interesting solution. I'd never think to use polynomials here.

@Bodyknock 3 жыл бұрын

13:25 I’m sure it’s obvious but I’m missing the reasoning why (2a + 5)^2 is either 25 or 9? What ruled out other possible squares?

@shohamsen8986 3 жыл бұрын

Me too. I dont follow the logic of difference between two consecutive square is odd. Anyway,my guess is we are looking for pythagorean triples and 25 is the only one that works.

@shohamsen8986 3 жыл бұрын

It would be nice to see a proof though.

@michielhorikx9863 3 жыл бұрын

You can simply write out (n+1)^2 - n^2 and notice that's 2n + 1.

@makotoniijima862 3 жыл бұрын

@@shohamsen8986 This is always true. Proof: For two consecutive numbers p,q∈N where q=p+1, it follows that (p^2−q^2)=(p−q)(p+q). Moreover, (p^2−q^2)=(p−(p+1))(p+(p+1)) which gives (p2−q2)=−1(2p+1) Hence the number is in fact odd. QED I didn't prove the other case which is q = p-1, because if your swap p and q the proof stills holds. Another way to prove this fact is very simple: Consecutive numbers have different parities and squaring preserves parity. The difference of numbers with different parities is odd.

@DylanNelsonSA 3 жыл бұрын

It's also important that the difference between squares increases as the squares become larger, so that once you have two consecutive squares that differ by 9, there can't be two larger squares (consecutive or not) that differ by 9. But if you prefer, you can also solve the equation x^2 - 9 = y^2 by factorising it as (x - y)(x + y) = 9, and then considering the cases that you get by letting x - y be each of the factors of 9. We can eliminate half of the work by assuming that x and y are positive so that x - y is the smaller factor, and so that both factors are positive. This allows us to identify what the squares themselves are, but then once we know that x^2 = 25, we have to be careful to consider both the possibility that e.g. 2a + 5 = 5, and 2a + 5 = -5. (i.e. I'm not saying that we can assume that 2a + 5 itself is positive; I'm saying that we should substitute x = |2a + 5| and y = |m| to reduce the number of cases.)

@Tanya-ol6cc 3 жыл бұрын

12:38

@cbbuntz 3 жыл бұрын

I need more videos about Chebyshev and Jacobi polynomials in general. There are so many cool and useful properties involving Chebyshev polynomials and Jacobi polynomials are a really beautiful piece of math that I don't see many people talking about.

@picklesauce7983 3 жыл бұрын

WLOG a>b and bounding the squares gives a simple solution

@miruten4628 3 жыл бұрын

Let A = alpha, B = beta. A = -2 => B = -2 => (a, b) = (4, -4) doesn't work. A = 1 => a^2 + 4b = (B - 1)^2 and b^2 + 4a = B^2 - 4B - 4. Second one is equal to a perfect square x^2 iff x^2 + 8 is a perfect square y^2 iff 8 = (y + x)(y - x), which gives the two cases B = 5 or B = -1 in the end. B = -1 gives (a, b) = (0, 1), which we ignore. B = 5 gives the sporadic solution (a, b) = (-6, -5). This case was the most interesting! All in all: (a, b) or (b, a) = (-4, -4), (-6, -5), (m^2, 0) or (n+1, -n), where m ≥ 0, n > 0 are integers.

@megauser8512 3 жыл бұрын

Nice!

@shreyamjha3058 3 жыл бұрын

It was "nice"to be able to visualise that they were actually discriminants of some quadratics which could be used ,a great soln

@IanXMiller 3 жыл бұрын

Only other solutions I got were the pairs (-6,-5) and (-5,-6).

@khoozu7802 3 жыл бұрын

Oh really? How about (3,-2)

@IanXMiller 3 жыл бұрын

@@khoozu7802 Michael described that one when he wrote the solution (t,1-t). You have listed the t=3 case.

@srikanthtupurani6316 3 жыл бұрын

Can you make some videos on combinatorial number theory connected to imo. They involve very beautiful ideas.

@anasuit1111 3 жыл бұрын

this is a perfect solution. I couldn’t solve this problem.

@luffnis 3 жыл бұрын

Can you show all cases im to lazy😩

@shohamsen8986 3 жыл бұрын

Dude use your ultra instinct to solve the problem

@GraemeMulvaney 3 жыл бұрын

I like how we get homework now

@AKhoja 3 жыл бұрын

Why must it be the case that BOTH quadratics have two integer solutions? Can someone clarify why can't one of the discriminants be 0?

@AM-bs1iu 3 жыл бұрын

Because in that case a and b must be equal to 0

@japanada11 3 жыл бұрын

That was a mistake: the discriminant can be zero, actually! His solution never actually uses the fact that the two roots are distinct; in fact, the solution (a,b)=(-4,-4) at 14:30 corresponds to alpha=beta=2.