Belgium-Flanders Mathematical Olympiad | 2005 Final #4
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- Опубликовано: 30 июл 2024
- We present a solution to final problem 4 from the 2005 Belgium-Flanders Mathematical Olympiad.
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Trying your problem now!! Awesome video as always!
504^2 and 96^2.
I got worried a bit since 2020 had quite a few factors. But I soon noticed (odd factor)*(even factor) wouldn’t work.
Thank you for promoting this amazing channel.
Hii blackpenredpen 💓💓💓
ty bprp for promoting one of my favourite channels and luv ur content too :D
Mis dos héroes!!
Finally - an Olympiad problem that didn't make me feel inadequate. :)
Same here :)
you must notice that this problem is for 12 to 14y.o. (very talented) kids, passed this thing, never reached final, very hard thing
@@francoisSwine you are totally right.
@@stevenrectangle399 What? Besides the number of hours they spend, they surely have some natural ability to be more adept to mathematics than the average youth.
@@stevenrectangle399 jealous
The first part of the solution is unnecessary I think. The second part gives you the two solutions and, at the same time, proves that those are the only solution.
Well it is unnecessary in a "refined" solution, but it is a good step to get to the final result if you do it yourself.
Yeah, it is unnecessary, but if you don't know what the result will be, it is a good first step to take.
in a timed exam like this, least work is favorable, but you wouldn't know that the first solution was unnecessary to begin with
I like the second part more(as is the more general approach), but isn't the first enough, so the 2nd part could be counted unnecessary? If n,m are natural numbers and non perfect squares at the same time then sqrt(n) + sqrt(m) is not a natural number itself. I think I managed to prove it, but might be wrong...
@@basicallyph0r i disagree, you spend the most time finding a direction. You have widely enough time to actually write. You need to speed up the time it takes to figure out what to do. I dont know why its in some ppl's nature to always try and find an argument against comfortable methodology.
Here is a simple alternative proof:
if sqrt(n+2005) + sqrt(n) = k then
sqrt(2005) - sqrt(n) = p/q
(multiply by the conjugate)
Therefore sqrt(n) and sqrt(n+2005) -linear combination- are rational.
Therefore sqrt(n) and sqrt(n+2005) are perfect square.
Hence the solutions.
could you explain why sqrt(n+2005) + sqrt(n) = k implies sqrt(2005) - sqrt(n) = p/q? doesn’t multiplying by conjugate give 2005 = k(sqrt(n+2005) - sqrt(n))?
@@aaravbajaj1909 Yeah, so sqrt(n+2005)-sqrt(n) = 2005/k and is hence rational
sqrt(n+2005)-sqrt(n) = 2005/k together with sqrt(n+2005) + sqrt(n) = k give you that both sqrt(n+2005) and sqrt(n) are rational numbers
Did the same
But the conjugate is just sqrt of n minus sqrt of n + 2005 and when you multiply that by the original you just get 2005 so i don't see how that helps.
I think you don't need 2nd part at all:
Suppose k is integer and k=sqrt(n)+sqrt(n+2005)
Than sqrt(n+2005)=sqrt(n)-k
Square both sides: n+2005=n-2k*sqrt(n)+k^2
Since k and n are integers you can tell that sqrt(n) is also an integer so n is a perfect square
Same work you can do with sqrt(n+2005)
In fact for all integers a and b if sqrt(a)+sqrt(b) is an integer than a, b must be perfect squares
Once you've determined sqrt(n) to be an integer, it immediately follows that sqrt(n+2005) is too, since their sum is an integer.
Then how will u determine values by skipping second step
@@fearlessprabal2996 in the second step he didn't determinate any value
Here you are actually relying on a fact that if sqrt(n) is not an integer then it is an irrational number. It is indeed true, but needed to be at least stated or even proved
I am so proud of myself for solving it before watching the video
Good
A clever trick is to recognize that the equation we are solving for is the quadratic formula for the positive root of the polynomial equation x^2 -2sqrt(n)x-2005=0. Since we are lookiing for integral roots we should look to factor the above polynomial with the factors of 2005 which happen to be 1 and 2005 and another one is 5 and 401. From which we can solve for n i.e 2*sqrt(n)=2005-1 and 2*sqrt(n)=401-5. And there are no other solutions since otherwise they would not be natural numbers.
How is that a quadratic i dobt see the equation youbare referring to?
@@leif1075 √n+√(n+2005)=m
m - √(n+2005) = √n
m^2 - 2√(n+2005)m + n + 2005 = n (square both sides)
m^2 - 2√(n+2005)m + 2005 = 0
Want to solve this equation for the positive root, m.
First method was what came to me to my mind as soon as I saw the problem. 2nd method is more refined and required more thinking from my end. Keep the problems coming. Greetings from Germany!!
I honestly don’t know how I found this channel but I’m glad I did
Great video, great channel. As a french, I find the way you explain the problems easy to understand
Such an elegant solution! Excellent work as always.
Great math channel - Subscribed! Math is best learned by solving problems.
I love this channel. Its literaly like having classes
Class on steroid you mean. I wish I had problems like that to solve when I was at the uni
You are very underrated. I love your videos
The deduction “X^2 is natural then X is natural” is wrong. See 6:48. However, by using the assumption m is natural, you may deduce 2005^2/m is natural, and of course 2005^2 has the same factors as 2005 so the rest of the solution can be modified accordingly.
it must be natural because it should be divisible by 2
Your comment is wrong. Of course 2005^2 does not have the same divisors as 2005. However, we do have that (m-2005/m)^2 is natural, implying that the expression m^2+2005^2/m^2-4010 is natural. The fact that m is natural implies m^2 is natural, and clearly 4010 is natural, thus m^2 divides 2005^2. Then you can get the solution, because m^2 | 2005^2 implies m | 2005 given m is natural.
Me: Who's a good boy?
My dog: If n and 2005 + n are both squares, then we are the good boy.
I did this and its wayyyyy easier.
n^2 and (n+1)^2 differ by 2n+1, or an odd number. You can find solutions by looking for sums of consecutive odd numbers that add up to 2005, 2020, or whatever integer you add to n.
Thank you, I love this question and also your solution.
last part, don't even need to plug and check. Since n is a natural number, the radical sqrt(n+2005) rules out m=1,5 immediately since all the radicals are positive and sqrt(2005)>5. We therefore have 2 m's left, corresponding to the two solutions.
Solution for 2020 gives 504^2 and 96^2. Keep it up!
Alessandro Papi daje teniamo alto il lavoro italiano
Ottengo anche 501^2/4 e 81^2/4
@@Neiltwiss 501^2/4 and 81^2/4 are not natural numbers!
Ah, I wish I had these when I was a kid. Great video
So the trick mentioned obliquely in the video is that the (m - 2020/m) has to be even so that the 1/4 goes away. But m and 2020/m are just the pairs of factors of 2020, so you need pairs where both are even or both odd. Since 2020 is even, there are no factor pairs that are both odd, so that leaves just 2*1010 and 10*202, and thus you have n=504² or n=96².
Since sqrt(n) and sqrt(n+2005) both need to be in Z, the second step can be skipped.
Simply observe that after sqrt(n)=1002, the difference in squares of consecutive integers becomes larger than 2005. Example: 1004^2-1003^2=2007>2005.
If 2005 is replaced by 2020, then the method exposed here leads to two solutions n=504^2 and n=96^2.
If 2005 is replaced by 2020, then 2005 is going to suck.
@@terracottapie bruh
@@terracottapie nice
Thank you for verifying my answer!
Well done, Amazing video as always.
Very clear explanation. Thanks.
I approached it in this way:
The RHS is in N, so the LHS should be in N, n should be a perfect square, say z^2, with z in N. Then the second term can be separately calculated as z^2+2005 = y^2 where y is in N. Rearranging, we get
2005 = (y-z)(y+z)
Factorizing 2005 as 5×401 and 1×2005, we get to solve two pairs of equations for z.
Simple solution of a complex looking problem. Awesome...
When you have sqrt(n) and when you square, a simple way will be this:
n=(m^4-4010m^2+2005^2)/4m^2, so for n to be natural, 4 need to divide the numerator. If we take mod 4 from numerator, we will have: m^4-4010m^2 ≅ 3 mod 4 (because 2005^2 end on a 5) so we basically have m^4 ≅ 3 (mod 4) which is impossible.
(act like ≅ means congruent to)
More thorough than my solution. I figured n must be a perfect square of a rational number so n = (a/b)^2. Then working on the second root we can take out (1/b) as a factor and square what's left to get a^2 + 2005b^2 = c^2 for some integer c.
Then 2005b^2 = (c+a)(c-a).
At this point I gave up with the rational root and went full integer with b=1. It then remains to check each pair of factors of 2005 against the RHS. eg
2005 = c+a
1 = c-a
This gives a = 1002.
Factors of 401 and 5 give a=198.
It is so easy when you explain it!
Magnifico video. Congratulations.
Very good video . Both the methods are interesting
LOVE YOU PROFESSOR!!! also,great number theory problem.
Can you do a video on how to prepare for USAMO,IMO? (or putnam?)
Thanks a lot
The 2nd part is neat, but I suspected that for any natural a, b, for √a+√b to be natural, a & b must both be natural, and I proved this. So part 1 is now enough to solve this problem and any variant, such as replacing 2005 by 2020 as proposed.
My proof (which may not be the most elegant) is as follows:
Suppose a, b and √a+√b=c are positive natural numbers (the result is trivial if a or b =0)
Then c^2=a+2√(ab)+b,
So √(ab)=m is rational, so natural [1st result needed from elementary number theory].
Suppose k=gcd(a, b), a=kp, b=kq, so that p, q are natural and coprime.
Then √(pq)=m/k is rational, so natural, and as p, q are coprime they are both perfect squares [2nd result needed from elementary number theory], √p=s, √q=t say, and √a=s√k, √b=t√k.
So c=√a+√b=√k(s+t). So √k is rational and hence natural, so a & b are both perfect squares QED
In the first line you said: "a & b must both be natural" but I suppose you mean that a & b must both be perfect squares or you could say √a & √b must be natural. Very nice proof by the way.
Excellent !!
Yes, very good
Is it just me or is he done at 6:03 since right hand side is rational and left hand side must be irrational. The left hand side is irrational because if it was rational n but be a perfect square and the assumption is that it's not.
I feel like I'm getting better in this kind of stuff. I used to fail miserably but now I only need a few hints.
User Name how’d you get to that point? Was it learning more maths or attempting more of these problems?
@@daedyg793 both honestly. But at a certain point, I feel like that I would stop needing to learn more math to get better at problem-solving.
We can actually extend this problem. It is well-known that if sqrt(a_1) + ... + Sqrt(a_n) is rational, then each a_i is a square of rational (This can be proved by using 2^n conjugates of this expression and by considering polynomial with roots of such conjugates ). So if one plays around with constants, problem could get really interesting. If I am not mistaken, there is even a generalization of presented fact : sum of a_i raised to power the b_i is rational, iff a_i^b_i is ratjonal ( all numbers used here are rational).
Excellent work
Very well done! Say, starting around 6:07 you suggest "taking the half out of it" and proceed to do so, but I would think we'd have (m^2 - 2005/m) instead of (m - 2005/m). In other words, we are still left with m^2 on the left after we take out the 1/2?
There is a slight conceptual error in the end. m = 1 and m = 5 are not possible results given that, since n is natural, m >= 1 + sqrt ( 2006 ) > 45 therefore m has to be larger than 45 so the first two solutions are invalid. These appeared as spurious roots after squaring the equality.
It's worth explaining that the orange-box-is-natural to red-box-is-natural is because there exist no non-integer rational roots of integers. You can get it from the residues of the exponents in the prime factorizations -- if it's not an integer result, they don't match.
Nice! But you dont need suppose m=1 and m=5 because m-2005/m must be natural. So m=401 or m=2005 is the only possibilites.
Am i wrong?
Good Job. Bye!
Equaciona Com Paulo Pereira I believe (m-2005/m)^2 is what needs to be natural. I have a feeling though m=1 or 5 aren’t possible solutions because the original equation (n)^1/2+(n+2005)^1/2>2005^1/2>5 or 1. They appear to give a valid n because if you consider the negative root of n^1/2 or (n+2005)^1/2 then you can get m being 1 or 5 :)
@@joshyman221 I agree with you. At first glance, I understood that he supposes that m-2005/m is natural. But we could not assume that since (m-2005 / m) ^ 2 is natural. I'll see again. Thanks! Hugs.
He said "at least it has to be natural" (for m-m/2005) what are the other conditions that he didnt mentioned(sorry for my english by the way )
Poxa, estou surpresa de te ver aqui, alias, sou sua inscrita.
Brilliant question and explaination sir 👌👌👌👌👌
8:04 for the red "thing" to be a natural number, m and (2005/m) doesn't have to be a natural number. For example:
m - 2005/m = 1 -> natural number
m -1 -2005/m = 0
m^2 -m -2005 = 0
solving quadratic equation -> m = 1/2 +- square_root(8021)/2 -> real number, so (2005/m) is also a real number
Yeah that 2005/m has to be natural for (m - 2005/m) to be natural didn't seem trivial too me. Thing is that we have postulated that m is natural and that does mean that in this context 2005/m must be natural for the whole to be natural.
I love your videos.
Woah, great video. The last problem give me: n=96^2 and n=504^2 as only solutions
Oh good, I am right :)
same solution (as in the video), and same result as yours :)
samee, yay i understood :)
Yeah, i tried it too, same results!!
@HERMAN LEONG XIN YANG - No, it’s easier than that! (m - 2020/m) must be even, so only factor pairs of 2020 that are both even will work. Thus, you only need to consider factor pairs of 2020/4=505. The only possibilities there are 1•505 and 5•101, which leads to the two solutions given. :-D
I like all your solutions
You are smart guy. Keep the good job brother.
It's now 4AM at my place and I have a business ethic exam at 9AM but I still have no idea why I'm watching this until the end.
I have a question: aren't we supposed to say that m-rad(n) is higher or equal to zero and how could we prove it? I'm referring to the equation that you wrote at 5:06.
Anyway excellent explanation, all clear!
There is one general fact that was used in one recent russian olympiad of informatics:
In fact, the only integer solutions of sqrt(A)+sqrt(B)=sqrt(C) is in form x * sqrt(d) + y*sqrt(d) = z*sqrt(d) where d is square free number. Proof is simple: square both sides, you'll get A^2+2sqrt(AB)+B^2=C^2, this means 2sqrt(AB) is integer.
lets say A is x^2*d where d is square free. And B is y^2*e, then 2sqrt(AB) is 2*sqrt(x^2*d*y^2*e) = 2*x*y*sqrt(e*d).
Square free e and d means that any prime in factorization is only once. But if p is in factorization of e and not in factorization of d, then e*d is not square, thus, for each p either both has p in factorization, or both don't have it in factorization. Therefore d and e are equal. Now we have proof that it's in form x * sqrt(d) + y*sqrt(d) = z*sqrt(?). Obviously x*sqrt(d)+y*sqrt(d) = (x+y)*sqrt(d).
Similar fact holds about sqrt of any other power. In other words, all integer solutions of A^(1/n) + B^(1/n) = C^(1/n) in form of x * (d^(1/n)) + y * (d^(1/n)) = (x+y)*(d^(1/n))
But this fact require some very advanced math. I don't know easy proof. Also, I don't remember name of this fact.
Nice, this is similar to the proof I gave just now that that for any natural a, b, for √a+√b to be natural, a & b must both be natural in my comment above. But someone else gave a much simpler proof using the difference of two squares, (√a+√b)(√a-√b)=a-b.
For \sqrt(n)+\sqrt(n+2020) to be an integer, n must be 96^2 or 504^2
I solved it by using the fact that all perfect squares k^2 can be written as a sum of first k odd numbers, or as (k-1)^2+(2k-1).
So, if we want n+2005 to be a perfect square, we just need to find the ways to write number 2005 as a sum of consecutive odd natural numbers, and we can do that by finding the factors of 2005 that are smaller than sqrt(2005). We found 1 and 5, which means that it can be represented as one number: 2005, or as a sum of five numbers: 397+399+401+403+405.
Therefore, we can conclude that for n+2005 to be a perfect square, (2k-1) needs to be 397 or 2005, so k=198, 1002, and n=k^2=39204 and 1004004.
Good job ! Thanks.
Very nice! I like it!
Thank you really nice explained .
Nice problem and explanation, but can be done but more easily....
Since (√[2005+n] + √n)(√[2005+n] - √n) = 2005,
and (√[2005+n] + √n) is a natural number say k, (√[2005+n] - √n) should be a positive rational number p/q, so k*p/q = 2005, (p, q positive integers, but we shall show q = 1)
Subtracting (√[2005+n] - √n) from (√[2005+n] + √n) gives
2*√n = k - p/q,
but since right hand side is rational, left hand side can't be irrational and n has to be perfect square, and since left hand side is now natural number, right side should be too, which means q = 1 (and p = 2005/k)
Also as left side is even, means k and p should be both factors of 2005 with same parity i.e, odd as 2005 is odd,
and as p < k, the only sols for (p,k) are ordered pairs (1, 2005) and (5, 401), giving two values for n
As for your second problem with 2020, we should now look at (p, k) to be factors of 2020 with same parity.... Since we can't have an even and an odd factor, and luckily 4 is a factor (2020 = 4*505), we can distribute a 2 to each of factors of 505.... Hence (p,k) for p < k are (2, 1010) and (10, 202)
giving 2 values of n, 504^2 and 96^2
Thanks for the explanation, but i cant seem to understand how you got the n after getting the ordered pair.
Did you square the equations?
@@nahfid2003
If you followed till this step
2*√n = k - p/q,
and why n is perfect square and q is 1, after finding ordered pairs (k, p)
it naturally follows from squaring both sides that 4*n = (k - p)^2
you have to make a channel in which you solve math problems
@@fakeit6339
Thank you! ☺️🙏
The simplest way is the no. Is 2005 so reduce 1 and make half it gives us 1002 but only one root we can get though this.
8:35 seems a bit off to me since plugging in m=1 into the original equation produces "no solution"
Edit: same with m=5 but as he mentioned, he other values of m are the "correct" values which produce the same n
There is a little problem in this otherwise gr8 solution. m=1 does not give n=1002^2. It's obvious that the 1st equation where we assume m tells us that its greater than sqrt(2005)
If k is a natural number, then sqrt(n) + sqrt(n+k) = m for some natural numbers n and m iff m divides k and m and k/m have the same parity. In this case, n is given by 1/4 (m-k/m)^2. k=2020 yields the solutions [in the form (m,n)] (1010, 504^2) and (202, 96^2).
Found the same thing. To have solution to this problem, k has to be a multiple of 4 or an odd number. It also show that you can't find two perfect square whose difference is any 2x(any odd number)... If and only if i'm not rusty :D
At 7:48 sounds like `in order to be x^2 a natural number, then x has to be a natural number` which does not seem correct. Opposite example: (sqr(2)) ^2 is natural, but sqr(2) - isn't.
You are correct, the solution is correct but the path is wrong. For the orange box to be natural number you have to expand it and find out 2005^2/m^2 has to be N which is the same result.
Perfeito. Obrigado professor 😃
Proud to say I got this one in my head, only because 2005 factors pretty easily 😅. That said, the vast majority of these videos aren’t anywhere near as simple! Love these videos, glad RUclips put them in my feed!
10:30 In a way, it's wrong. Given how you defined m as sqrt(n) + sqrt(n+2005), m clearly cannot be 1. You find back a solution anyway because those squaring of expressions some time earlier have added solutions that do not fit the original statement of m = sqrt(n) + sqrt(n+2005)
Wow: I actually understood this from start to finish
Thank u man you are great
Since 5:37 : More elegant idea 2005 = m*(m-2*sqrt(n)) and now m is natural number, so substitute m with dividers of 2005 ={1, 5, 401, 2005}
Nice problem I am solving after watching the videos
n = 254016 and n = 9216 by using the second method and exhausting all possible factors.
[8:27] Micheal states that m-2005/m is a natural number.
[8:46] In fact, 1-2005 is NOT a nat. number
[9:43] In fact, 5-401 is NOT a nat. number
I do agree that both m = 401 and 2005 are a more consistent way to reach the answer.
Yeah, he should've said (m-2005/m)^2 must be natural.
He should have said that ¼*(m-2005/m)² is a natural number only when m-2005/m=2k with k an integer and this is only possible when m is a factor of 2005, leading to m=1, m=5, m=401 and m = 2005.
When taking the squares in 5:11 he also should have mentioned the condition m>sqrt(n). That is why the solutions m=1 and m=5 are incorrect and the solutions m=401 and m=2005 are correct, leading to n=198² and n=1002² respectively.
Wooow, what a cool feelings! That's only one your video's problem s.t. i understood you
awesome video !!!!!
I wonder if this is possible...
Where
Sqrt (n) = a.56
And sqrt(n) = b.44
And sum of these will also be a natural number... But both terms are non natural individually....
For 2020 the ( ONLY ) solution is
506^2-2020 ,504^2 &
106^2-2020 ,96^2
Another method:
For √n+√n+2005 to be a natural number n+2005 and n have to be a perfect square
Let m^2=n, n+2005=i^2
Hence:
m^2+2005=i^2
2005=i^2-m^2
2005=(i+m)(i-m)
factors of 2005 are 1,2005,401 and 5
Checking for 401 and 5 pair
(401)(5)=(i+m)(i-m)
401=i+m.......equation1
5=i-m.........equation2
On solving we get m=198 and i=203
Pause at 6:16, quick way to argue, if n is not perfect square, then LHS is irrational, but RHS is rational, contradiction. Therefore, n must be perfect square. Again, we can do the similar way to show n+2005 must be perfect square.
I have no idea how he got m(sq) - 2m sqroot n for the 2nd solution. Where did the 2m came from? Sorry... I'm pretty confused
Wade's Introduction to Analysis offers a method where 2005 is replaced with a prime number that uses real analysis, in particular, the least upper bound property of the reals. I didn't do it that way. Anyone know how?
For the question at the end the answer is n=504^2= 254016 and also n=96^2 = 9216
Yeah I got the same values.
If (m-2005/m)² is an integer , how do we know that (m-2005/m) is also an integer? Irrational numbers can square to a rational number
Got a question for you. Are you assuming all square roots are positive? If not, you can probably use x-y as the larger one, and vice versa. I believe that that would produce a different solution that is also correct.
The square root symbol just means the positive square root. If the negative square root were allowed, there would be a +/- sign in front of it. However, in either case, n must be a square number, so while the value would be different in the end, n would be the same
I am not sure about this, but isn't the first method fullproof, so is there any need for the second part (from 4:45) to be carried out?
Please help me, why or why not.
Thanks
Remark: m - 2005/m must be integer, not only natural
@Adam Morris I think they meant that Michael saying m - 2005/m must be natural is incorrect since it could also negative and its square would be natural.
I believe the step at 6:50 is not necessarily going to give you all the solutions.
Starting with n = 1/4 (m-2005/m)^2 being a natural number, (m-2005/m)^2 needs to be a multiple of 4, therefore (m-2005/m) must be of the form q=2*sqrt(k) where k is some positive integer.
In the video, it is assumed that q is an integer, which forces k to be a perfect square, however this is not guaranteed by the step that told us the form of q. Because k simply needs to be some positive integer, not necessarily a perfect square, this opens up the possibility for values of k which are not perfect squares and these could give you more solutions for m.
Now I've not done the calculation over with this step taken into consideration so I don't know if you end up with more solutions or not, but I think this would be a line of reasoning that would need to be taken into account in order to prove that you have all the solutions.
Still thanks for making the video but I feel jumps like these are what would trip up a student if they were to come across this situation somewhere.
At 9:53 How does 198 x 2 remove the denominator? Do we not need a 4 in the numerator for it to work out the way it did?
396²=(198×2)² take out the 2², it will get u four to cancel out the denominator
Nice. The problem looks harder than it actually is.
is the set of solutions infinite when you set the restriction of n to the quotients?
Thank you
Thanks!
At 6:13
m is a whole number, so (m^2-2005)/(2m) is rational. But when n is not a square, sqrt(n) is irrational. So the proof should already be done, right?
Hello.
Sqrt(n) is irrational only when n is a prime.
@@ramk4004 sqrt(8) is 2sqrt(2) and this is irrational thing but 8 is not a prime
Or we can say if m-2005/m is integer then it must be even. That implies (m-2005/m)/2 is integer, therefore n is a square of integer, which is what we wanted to show
@@ramk4004 nth-root of the number is irrational if the number isn't nth-power of another integer.
@@alexgan3219 I am sorry.
Both n and (n+2005) has to be perfect square ,
so let , n = m^2 and n+2005=p^2
then , m^2+ 2005 = p^2
then , 2005=p^2 - m^2 = (p-m)(p+m)= 1x2005 =5 x 401
this gives two sets of equations (p-m =1 & p+m = 2005 ) , (p-m =5 & p+m = 401 )
this gives m = 1002 , m =198
then n is 1002^2 , 198^2
Very good
9:15 we add 1 to negative 2005 and it becomes positive 2004? If its because that the square always yeilds a positive number, but the square was not still applied and is in fact a number above, so we need to keep that square and not apply it i think
(-2004)²=(-1.2004)²=(-1)².(2004)²=1.(2004)²=(2004)²
Or in general: (a-b)²=(b-a)² with a=1 and b=2005
(a-b)²=(-(-a+b))²=((-1).(-a+b))²=((-1).(b-a))²=(-1)².(b-a)²=1.(b-a)²=(b-a)²
(1-2005)²=(2005-1)²=2004²
(-2004)²=(0-2004)²=(2004-0)²=2004²
These olympiad problems are so good
Thanks sir
I used to participate in this competition when I was still in school and it was a load of fun.
1 question. How do we know that there is no decimal number m that divides 2005 in a way that m + 2005/m is a natural number?
Here m is the number sqrt(n)+sqrt(n+2005), which is assumed to be an integer
I solved in an other way.
take the conjugate. It's still a natural number. with an additional information that sqrt(n)-sqrt(2015) is an integer.
adding two integers yields an integer summing up sqrt(n)+sqrt(n+2015) and sqrt(n)-sqrt(n+2015) as well as taking the difference. now we've to the conclusion that sqrt(n) and sqrt(n+2015) are both integers thus n and n+2015 are perfect squares. And we carry on with the same approach.
Good day, sir !!!
thank you