Surely it is easier than this? Just note that (n²+1)² = n⁴ + 2n² + 1 < n⁴ + 3n² + 2 < n⁴ + 4n² + 4 = (n² + 2)². The given expression is strictly sandwiched between two consecutive squares, so cannot be one itself. Also the fourth (respectively second) powers of n are unnecessary: the same thing is true of all n² + 3n + 2 for the same reason.
Those powers are actually necessary. For m^2 + 3m + 2, it is true of all *positive* integers m that it cannot be a square. But if m = -1 or m = -2, then you get 0, which is a perfect square. But there is no real number (let alone integer) n such that n^2 = -1 or -2, which makes the original claim true.
But not as instructive. I appreciated this video so much, and if the video had been what you said in less than 30 seconds instead of this educator's witty insight into number theory, I would not have subscribed in the first minute!
Nicely done. Here is my proof: n⁴ + 3n² + 2 factors as (n² +1)(n² +2). Let n² +1=x. Then it becomes x(x+1), or the product of 2 consecutive integers. But what is x(x+1)? The formula for triangular numbers is well known, it is [x(x+1)]/2. So x(x+1) is simply twice a triangular number. Now consider the squares. Every square is the sum of 2 consecutive triangular numbers. List the triangular numbers: 0,1,3,6,10,15,21, etc. So 0+1 is the first square, or 1². The next square is 1+3, or 2². The next square is 3+6, or 3². Etc. Now consider again x(x+1) Since it is twice a triangular number, that means it is 0+0, 1+1, 3+3, 6+6, etc. But notice this pattern. We have already established that every square is the sum of 2 consecutive triangulars. With the exception of 0, any triangular number added to itself falls short of the very NEXT square. In other words, 1+1 cannot be a square since AT LEAST 3 has to be added to 1 to reach the next square, 2². 3+3 cannot be a square since AT LEAST 6 has to be added to 3 to reach the next square, 3². 6+6 cannot be a square since AT LEAST 10 has to be added to 6 to reach the next square, 4². So the pattern continues to infinity. One other case: 0. In this case, 0+0 IS a perfect square, but plugging it back into the original quartic, n⁴ + 3n² +2 yields 2, which is not a square. QED, proof complete.
Actually it is very easy to show that a product of 2 consecutive integers cannot be a perfect square. Just note that x(x+1)=x^2+x. Assuming that x>0 we have: x^2
My solution: n^4+3n^2+2=k^2 (n^2+1)(n^2+2)=k^2 (n^2+1) and (n^2+2) don't share any factors, since they differ only by 1, so they both have to be squares for (k) to be an integer so: (n^2+1)=a^2 (n^2+2)=b^2 then: b^2-a^2=1 (b-a)(a+b)=1 either: b-a=1 a+b=1 or: b-a=-1 a+b=-1 which makes a=0 and b=+-1 resulting in: n=sqrt(-1) which is not an integer.
Very good proof, nice and easy ! I found one which is longer (and less smart), but it works… there it is : Let X = n^4 + 3 n^2 + 2. Then 4X = 4 n^4 + 12 n^2 + 8, and 4X = (2 n^2 + 4) (2 n^2 + 2) = (2 n^2 + 3)^2 - 1. Then 4X can’t be a square (because it is a square minus 1), and X can’t be a square (because 4 is a square). Thanks for your videos 🙂
I have another solution: Assume that such a k as an integer exists. such that n^4 + 3n^2 + 2 = k^2 n^4 + 3n^2 + 2 is (n^2 + 1)(n^2 + 2), but we can write it as n^4 + 3n^2 + 2 = (n^2 + 1) (n^2 + 2) = (n^2 + 1.5 - 0.5) (n^2 + 1.5 + 0.5) = (n^2 + 1.5)^2 - 0.5^2 Then, (n^2 + 1.5)^2 - 0.5^2 = k^2 (n^2 + 1.5)^2 = k^2 + 0.5^2 (2n^2 + 3)^2 = 2k^2 + 1^2 This is a pythagorean triple. However, no such pythagorean triples exist with 1 as part of the primitive triple, other than +/-1, 0 and 1. k = 0, but this gives that 2n^2 + 3 = +/-1, which has no solution in integer n. So no such n exists. The reason as to why we seek for primitive triples rather than all triples is that for non-primitive triples, they can be multiplied by a constant to get the next triple. [specifcally k^2], so any equalities made are preserved. As a result, we just need to check for if it follows a primitive triple. === Another way is that if we tried to "force it" to make a triple of a,b,c we'd get (a*[2n^2 + 3])^2 = (2ak)^2 + a^2 a*(2n^2 + 3) = c and 2ak = b, so ideally 2n^2 + 3 = c/a and 2k = b/a. But in a primitive triple, a,b,c are coprime, so the fraction b/a and c/a cannot simplify to an integer. so in the first place the only possibility is that if if k was 0 and b was 0, but that leads us to the reasoning above. === Alternatively, we can just factor it as (2n^2 + 2k + 3)*(2n^2 - 2k + 3) = 1 Both these factors must equal each other and equal 1, since 1*1 = 1 and -1*-1 = 1. So this means k = 0, and 2n^2 + 3 = 1 or -1, but once again, same as above. === There's different reasonings as to how this works but I've listed all of them down.
Very easy visual proof as well, I shall attempt to explain in words: n⁴ + 3n² + 2 can be expressed as n²(n²+3) +2 You can now construct a perfect square that has lengths n². You have three n² by 1 lengths and 2unit lenths to add. Adding one [1*n²]lengths to the top/bottom and the other on the left/right. You nearly have a perfect square but are missing one piece. So you can add 1 to complete the square. At this point our equation would be n²(n²+2) + 1 is always a perfect square. But we still have n²+1 to add. At this point it should become visually apparent there is no way to add n²+1 to n²(n²+2)+1 to make a perfect square. You would always be missing at a minimum n²+2 more. (That is [n²+1]+[n²+2])
ATTEMPT: All square numbers are either 0 or 1 mod 4. Consider when n is even: n^2 is 0 mod 4. n^4 + 3n^2 + 2 = 2 mod 4, so the expression cannot be square. Now consider when n is odd: n^2 = 1 mod 4. n^4 + 3n^2 + 2 = 2 mod 4, so the expression cannot be square. By case analysis we are done.
3rd method: I assume there is an n such that n^4+3n^2+2 is a perfect square. Then there must exist an integer k for which n^4+3n^2+2=k^2. Let m=n^2. That means m>=0 and mis an integer. => m^2+3m+2=k^2 => m^2+3m+2-k^2=0 Quadratic formula m=-3/2+/-sqrt(9/4+k^2-2)=-3/2+/-sqrt(9+4k^2-8)/2 That means m can only be an integer if 9+4k^2-8 is a perfect square. 9+4k^2-8=(2k)^2+1. The only consecutive integers which are both perfect squares are 0 and 1. Therefor k=0. That means m=-3/2+/-1/2. However both solutions for m are negative and therefor n can not be an integer for k=0 as well. That means our assumption is wrong and there is no n for which n^4+3n^2+2 is an integer.
Sir... u r such a SUPERB math teacher! 😮 ive always been impressed by ALL math videos u've produced b4. U R amazing, man! Hats off to u! ❤ pls keep being such an outstanding inspiration to us all!!! (this is from an aeronautical engineer who's always been summa cum laude in ALL math subjects from first grade to PhD level! 😋😁) God bless!
n⁴+3n²+2=(n²+1)(n²+2) Suppose there exists a prime p s.t. p divides n²+1 and n²+2. Then p divides their difference (n²+2)-(n²+1)=1. Therefore no such p exists. It follows that n²+1 and n²+2 are both perfect squares. Call them a² and b² and suppose a and b are positive. (b-a)(b+a)= b²-a²= (n²+2)-(n²+1) =1. However a is not equal to b and a+b is atleast 3. This contradicts the fact that n⁴+3n²+2 can be a perfect square.
If n⁴+3n²+2 is a perfect square, it can be written as a²+2ab+b² for every integer in a there will be a possible value of b also integer. For a = n² we have (n²)²+2(n²)b+b² = n⁴+3n²+2 2n²b+b² = 3n²+2 2n²b-3n²=2-b² n²(2b-3) = 2-b² n² = (2-b²)/(2b-3) But n² = a (a+b)² = ((2-b²)/(2b-3)+b)² =(2-b²)²/(2b-3)² + 2(2-b²)/(2b-3)b+b² (4-4b²+b⁴)/(2b-3)² + 2b(2-b²)(2b-3)/(2b-3)²+b²(2b-3)²/(2b-3)² (4-4b²+b⁴+(4b-2b³)(2b-3) +(b(2b-3))²)/(2b-3)² (4-4b²+b⁴+8b²-12b-8b⁴+3b³+(4b⁴-12b³+9b)/(2b-3)² (-3b⁴-9b³+4b²-3b+4)/(2b-3)² This expression can't be simplified in terms of integers factors, what means that a that is n² or b are never both integers at same I was pretty sure I could get a solution by this way, but easier, but that was my best
Hello, I am following you from Iraq. Can you explain this question? It came in the monthly exam at our school. Prove z^2-1/z^2+1=itanø , z = cosø + isinø
n^4 + 2n^2 + 1 = (n^2+1)^2 < n^4 + 3n^2 + 2 < n^4 + 4n^2 + 4 = (n^2 + 2)^2 therefore, we will need a perfect square that is between two consecutive perfect squares.
n^2 and n^4 are always equivalent mod 4 (0^2=0^4=0, 1^2=1^4=1, 2^2=2^4=0, 3^2=3^4=1). Thus the expression is equivalent to n^2+3n^2+2 = 4n^2+2 = 2 mod 4. This also proves the evenness of the expression since any integer that is equivalent to 2 mod 4 is even.
If the product of 2 consecutive numbers is a square, since consecutive numbers are coprime, both numbers must be squares. But the only consecutive squares are 0 and 1. Even without being able to factor it, if k^2 = n^4 + 3n^2 + 2, 4k^2 = 4n^4 + 12n^2 + 8, (2n^2 + 3)^2 - (2k)^2 = 1.
The first thing that came to my mind is that the expression is also equal to (n^2+1)^2 + (n^2+1), ie it’s a perfect square plus the base of that perfect square , in the form m^2 + m But the next perfect square after m^2 is 2m+1 apart. There is no way m^2+m is a perfect square
Let us prove that there does not exist an m in Z satisfying the equation m^2 = n^4 + 3n^2 + 2. To do this, we will examine the equation modulo 3. If no solution m exists in Z/3Z, then no solution can exist in Z either. While the converse is not necessarily true, this approach suffices to establish our claim. First, recall that for the prime number 3, Wilson's theorem implies a^2 ≡ 1 mod 3. Using this, we reduce the given equation modulo 3. Since 3n^2 ≡ 0 mod 3, the equation becomes: m^2 ≡ n^4 + 2 mod 3. Noting that m^2 ≡ 1 mod 3, we rearrange to obtain: 1 ≡ n^4 + 2 mod 3, or equivalently, n^4 ≡ -1 mod 3. {At this point, we also note that Z/3Z is a finite field (a finite commutative ring with multiplicative inverses for all non-zero elements). Therefore, all arithmetic operations are well-defined and the equation we are considering is valid in this context.} Now, let us determine whether -1 is a quadratic residue modulo 3. Since 3 is a prime of the form 4k + 3, it follows that -1 is a quadratic non-residue modulo 3. This means there is no n in Z/3Z such that n^2 ≡ -1 mod 3. Consequently, it is impossible for n^4 ≡ -1 mod 3 to hold. Thus, we reach a contradiction: the equation m^2 = n^4 + 3n^2 + 2 has no solutions in Z/3Z, and therefore, it has no solutions in Z. This completes the proof.
I replaced n² with m and to make m² + 3m + 2 = k², and then solved for m² + 3m + 2 - k² = 0 via the quadratic equation. For k² and m are perfect squares and must both be positive integers. But the Q.E. gave me m = (-3 +/- sqrt(k² + 1)) / 2, which is k² is a square, forces m to be irrational or negative.
For all integer n, n² = 0 or 1 mod 4, thus n^4+3n²=n²(n²+3)=0 mod 4 thus n^4+3n²+2 = 2 mod 4 for all integer n. However, there is no square integer which is 2 modulo 4.
i think we can just end it on the first step the expression we have boils down to (n^2 + 1)(n^2 + 2) these are consecutive integers for all integral n, and the product of two consecutive integers can never be a perfect square, hence proven no?
Hi, since the comment section is the only way that outsiders can have a chance to contact you, would you please, consider putting a video about integrating probability functions? Currently I am having trouble integrating (f(x)/(1-F(x)))^0.5 over dx, where f(x)-pdf and F-cdf and the whole function is a conditional probability squared. Any useful hint would be appreciated!!
@ you did not put an email in the about section of your channel or in the video description, so I thought that only your patreons can directly contact you.
it’s kinda implied that m is non-negative because it is found between n^2 + 1 and n^2 + 2, and n^2 can never be negative, meaning that m can never be negative either.
Is my proof wrong, if so why? PLS HELP(im in grade 10) n^4 + 3n^2 + 2 (quadratic in terms of n^2) if it is a perfect square(assume), D= b^2 - 4ac = 0 (as roots are equal and real) D=9 - 4(1)(2)=0 D= 9 - 8 = 0 D = 1 =0 (contradiction) so original statement is wrong and it ain't a square number
Sir please do a limit question which was came in JEE Advanced 2014 shift-1 question number 57 it's a question of a limit lim as x-->1 [{-ax + Sin(x-1) + a}/{x + Sin(x-1) - 1}]^[{1-x}/{1-√x}] = 1/4 You have to find the greatest value of a It has 2 possible answers 0 and 2 But I want the reason that why should I reject 2 and accept 0 Because final answer is 0 Please help 😢
For all integers,prove that n^4+3n^2+2 is never a perfect square. First method Suppose there exist some integer m such that m^2=n^4+3n^2+2=(n^2+1)(n^2+2) x^2=4*9 x=±6
Do you think this solution would work? Suppose that n^4 + 3n^2 + 2 = k^2 for some k. Now we can show that because the factorisation of n^4 + 3n^2 + 2 is (n^2 + 1)(n^2 + 2), the expression minimally has a value of 2 at n = 0. [Or you can differentiate the expression, whichever feels more rigorous]. 2 >1^2 and 0^2, so this expression can never equal those squares. Now we consider 2 cases of k^2, where it is a prime square [so k is prime like 2,3,5,7,11], and k is a composite square, [so k is composite like 4,6,8,9,10] If k is a prime square, it's only factors are 1,k and k^2, so we could try to see if any values of n could make a value of k exist. Case 1: n^2 + 1 = 1, n^2 + 2 = k^2 Eq 1 shows that n = 0, so we'd get k^2 = 2 ,but k wouldn't be an integer as 2 isn't a perfect square. So this case is out. Case 2: n^2 + 1 = k, n^2 + 2 = k It is "trivial", but trying to solve the two equations yields 1 = 2, which means these 2 expressions can never equal each other at the same value of n. So the expression can never be a prime square. No prime values of k exist, so no n exists for which the expression is a perfect square. For non prime squares, its factors are 1,k and k^2, alongside any other factors between 1 and k, and k and k^2. We assume that the two factors n^2 + 2 and n^2 + 1 are factors that pair up to make a square that are not equal to each other [as we shown earlier]. So this means that: n^2 + 1 < k < n^2 + 2, where k is the non-prime factor of the non-prime square. This makes sense because factors will always pair up in a number. Since squares have 1 double counted factor, any factor pairs must lie on opposite sides of k. We run into another issue however. n^2 + 1 - (n^2 + 1) = 1, so these numbers are just 1 apart. There is no possible way to sandwich a value of k for any n, because these 2 are consecutive integers. So no value of non-prime k exists as well, so no n exists for which the expression is a perfect square. Since we showed that no value of k can exist for primes and non primes, as well as 0,1 this covers all the positive integers, and since k^2 = (-k^2), this covers the entire number range as well, so no value of k exists for which an integer n can produce a perfect square. As completion, Because the function is even, this also doubles as covering all integers, as f(n) = f(=n), so since we proved it for all positive n, any negative n will have the same result. Hence, no value of n can produce a perfect square. QED.
Sir,please solve this problem. The broblem is taken from Bdmo. 𝒇(𝒏) = 𝟏𝟎^𝒏 − (𝟓 + √𝟏𝟕)^n −(𝟓 − √𝟏𝟕)^n is a function which is valid for all integers 𝒏 greater than 𝟏. Prove that, 𝒇(𝒏) is always perfectly divisible by 𝟐^n+1.
Surely it is easier than this? Just note that (n²+1)² = n⁴ + 2n² + 1 < n⁴ + 3n² + 2 < n⁴ + 4n² + 4 = (n² + 2)². The given expression is strictly sandwiched between two consecutive squares, so cannot be one itself. Also the fourth (respectively second) powers of n are unnecessary: the same thing is true of all n² + 3n + 2 for the same reason.
Those powers are actually necessary. For m^2 + 3m + 2, it is true of all *positive* integers m that it cannot be a square. But if m = -1 or m = -2, then you get 0, which is a perfect square. But there is no real number (let alone integer) n such that n^2 = -1 or -2, which makes the original claim true.
But not as instructive. I appreciated this video so much, and if the video had been what you said in less than 30 seconds instead of this educator's witty insight into number theory, I would not have subscribed in the first minute!
Nicely done. Here is my proof:
n⁴ + 3n² + 2 factors as (n² +1)(n² +2). Let n² +1=x. Then it becomes x(x+1), or the product of 2 consecutive integers. But what is x(x+1)? The formula for triangular numbers is well known, it is [x(x+1)]/2. So x(x+1) is simply twice a triangular number. Now consider the squares. Every square is the sum of 2 consecutive triangular numbers. List the triangular numbers: 0,1,3,6,10,15,21,
etc. So 0+1 is the first square, or 1². The next square is 1+3, or 2². The next square is 3+6, or 3². Etc. Now consider again x(x+1) Since it is twice a triangular number, that means it is 0+0, 1+1, 3+3, 6+6, etc. But notice this pattern. We have already established that every square is the sum of 2 consecutive triangulars. With the exception of 0, any triangular number added to itself falls short of the very NEXT square. In other words, 1+1 cannot be a square since AT LEAST 3 has to be added to 1 to reach the next square, 2². 3+3 cannot be a square since AT LEAST 6 has to be added to 3 to reach the next square, 3². 6+6 cannot be a square since AT LEAST 10 has to be added to 6 to reach the next square, 4². So the pattern continues to infinity. One other case: 0. In this case, 0+0 IS a perfect square, but plugging it back into the original quartic, n⁴ + 3n² +2 yields 2, which is not a square. QED, proof complete.
Actually it is very easy to show that a product of 2 consecutive integers cannot be a perfect square. Just note that x(x+1)=x^2+x. Assuming that x>0 we have: x^2
I am very happy to find you on this platform! You are a very good teacher and friend. Thank you for your time.
My solution
n⁴ + 3n² + 2 = t²
Factoring:
(2n²+3)² = 4t² + 1
(2n²+3+2t)(2n²+3-2t) = 1
This implies t=0 and then:
2n² + 3 = 1
n² = -1, Absurd
My solution:
n^4+3n^2+2=k^2
(n^2+1)(n^2+2)=k^2
(n^2+1) and (n^2+2) don't share any factors, since they differ only by 1, so they both have to be squares for (k) to be an integer so:
(n^2+1)=a^2
(n^2+2)=b^2
then:
b^2-a^2=1
(b-a)(a+b)=1
either:
b-a=1
a+b=1
or:
b-a=-1
a+b=-1
which makes a=0 and b=+-1 resulting in:
n=sqrt(-1) which is not an integer.
Very good proof, nice and easy !
I found one which is longer (and less smart), but it works… there it is :
Let X = n^4 + 3 n^2 + 2. Then 4X = 4 n^4 + 12 n^2 + 8, and
4X = (2 n^2 + 4) (2 n^2 + 2) = (2 n^2 + 3)^2 - 1.
Then 4X can’t be a square (because it is a square minus 1), and X can’t be a square (because 4 is a square).
Thanks for your videos 🙂
Thank you for leaving in the "typos." 😊 It makes you a real person.
Once again an Absolutely beautiful proof. Love number theory.
Trying n = 0, 1, 2, -1, we find that n⁴ + 3n² + 2 ≡ 2 (mod 4), for all n. Similarly,
n² ≡ 0 or 1 (mod 4), for all n.
I have another solution:
Assume that such a k as an integer exists. such that n^4 + 3n^2 + 2 = k^2
n^4 + 3n^2 + 2 is (n^2 + 1)(n^2 + 2), but we can write it as
n^4 + 3n^2 + 2 = (n^2 + 1) (n^2 + 2)
= (n^2 + 1.5 - 0.5) (n^2 + 1.5 + 0.5)
= (n^2 + 1.5)^2 - 0.5^2
Then,
(n^2 + 1.5)^2 - 0.5^2 = k^2
(n^2 + 1.5)^2 = k^2 + 0.5^2
(2n^2 + 3)^2 = 2k^2 + 1^2
This is a pythagorean triple. However, no such pythagorean triples exist with 1 as part of the primitive triple, other than +/-1, 0 and 1.
k = 0, but this gives that 2n^2 + 3 = +/-1, which has no solution in integer n. So no such n exists.
The reason as to why we seek for primitive triples rather than all triples is that for non-primitive triples, they can be multiplied by a constant to get the next triple. [specifcally k^2], so any equalities made are preserved. As a result, we just need to check for if it follows a primitive triple.
===
Another way is that if we tried to "force it" to make a triple of a,b,c we'd get
(a*[2n^2 + 3])^2 = (2ak)^2 + a^2
a*(2n^2 + 3) = c and 2ak = b, so ideally 2n^2 + 3 = c/a and 2k = b/a. But in a primitive triple, a,b,c are coprime, so the fraction b/a and c/a cannot simplify to an integer. so in the first place the only possibility is that if if k was 0 and b was 0, but that leads us to the reasoning above.
===
Alternatively, we can just factor it as
(2n^2 + 2k + 3)*(2n^2 - 2k + 3) = 1
Both these factors must equal each other and equal 1, since 1*1 = 1 and -1*-1 = 1.
So this means k = 0, and 2n^2 + 3 = 1 or -1, but once again, same as above.
===
There's different reasonings as to how this works but I've listed all of them down.
Very easy visual proof as well, I shall attempt to explain in words:
n⁴ + 3n² + 2 can be expressed as n²(n²+3) +2
You can now construct a perfect square that has lengths n². You have three n² by 1 lengths and 2unit lenths to add. Adding one [1*n²]lengths to the top/bottom and the other on the left/right. You nearly have a perfect square but are missing one piece. So you can add 1 to complete the square.
At this point our equation would be n²(n²+2) + 1 is always a perfect square. But we still have n²+1 to add.
At this point it should become visually apparent there is no way to add n²+1 to n²(n²+2)+1 to make a perfect square. You would always be missing at a minimum n²+2 more. (That is [n²+1]+[n²+2])
ATTEMPT:
All square numbers are either 0 or 1 mod 4.
Consider when n is even: n^2 is 0 mod 4.
n^4 + 3n^2 + 2 = 2 mod 4, so the expression cannot be square.
Now consider when n is odd: n^2 = 1 mod 4.
n^4 + 3n^2 + 2 = 2 mod 4, so the expression cannot be square.
By case analysis we are done.
Never stop learning those who stopped learning have stopped living.
Nice solutions! I solved it by using congruence modulo 4.
3rd method: I assume there is an n such that n^4+3n^2+2 is a perfect square. Then there must exist an integer k for which n^4+3n^2+2=k^2. Let m=n^2. That means m>=0 and mis an integer.
=> m^2+3m+2=k^2 => m^2+3m+2-k^2=0
Quadratic formula m=-3/2+/-sqrt(9/4+k^2-2)=-3/2+/-sqrt(9+4k^2-8)/2
That means m can only be an integer if 9+4k^2-8 is a perfect square.
9+4k^2-8=(2k)^2+1. The only consecutive integers which are both perfect squares are 0 and 1. Therefor k=0. That means m=-3/2+/-1/2. However both solutions for m are negative and therefor n can not be an integer for k=0 as well. That means our assumption is wrong and there is no n for which n^4+3n^2+2 is an integer.
When a perfect square is divided by 4, the remainder is either 0 or 1.
0^2 ~ 0 (mod 4)
1^2 ~ 1 (mod4)
2^2 ~ 0 (mod4)
3^2 ~ 1 (mod4)
Now:
n^4 + 3 n^2 + 2 (mod4)
(n^2)^2 + 3 n^2 + 2 (mod4)
When the remainder is 0:
0^2 + 3 × 0 + 2 ~ 2 (mod4)
When the remainder is 1:
1^2 + 3 × 1 + 2 ~ 6 ~ 2 (mod4)
Hence, n^4 + 3 n^2 + 2 is never a perfect square.
Sir... u r such a SUPERB math teacher! 😮 ive always been impressed by ALL math videos u've produced b4. U R amazing, man! Hats off to u! ❤ pls keep being such an outstanding inspiration to us all!!! (this is from an aeronautical engineer who's always been summa cum laude in ALL math subjects from first grade to PhD level! 😋😁) God bless!
n⁴+3n²+2=(n²+1)(n²+2)
Suppose there exists a prime p s.t. p divides n²+1 and n²+2. Then p divides their difference (n²+2)-(n²+1)=1. Therefore no such p exists. It follows that n²+1 and n²+2 are both perfect squares. Call them a² and b² and suppose a and b are positive.
(b-a)(b+a)= b²-a²= (n²+2)-(n²+1) =1.
However a is not equal to b and a+b is atleast 3. This contradicts the fact that n⁴+3n²+2 can be a perfect square.
If n⁴+3n²+2 is a perfect square, it can be written as a²+2ab+b² for every integer in a there will be a possible value of b also integer.
For a = n² we have
(n²)²+2(n²)b+b² = n⁴+3n²+2
2n²b+b² = 3n²+2
2n²b-3n²=2-b²
n²(2b-3) = 2-b²
n² = (2-b²)/(2b-3)
But n² = a
(a+b)² = ((2-b²)/(2b-3)+b)²
=(2-b²)²/(2b-3)² + 2(2-b²)/(2b-3)b+b²
(4-4b²+b⁴)/(2b-3)² + 2b(2-b²)(2b-3)/(2b-3)²+b²(2b-3)²/(2b-3)²
(4-4b²+b⁴+(4b-2b³)(2b-3) +(b(2b-3))²)/(2b-3)²
(4-4b²+b⁴+8b²-12b-8b⁴+3b³+(4b⁴-12b³+9b)/(2b-3)²
(-3b⁴-9b³+4b²-3b+4)/(2b-3)²
This expression can't be simplified in terms of integers factors, what means that a that is n² or b are never both integers at same
I was pretty sure I could get a solution by this way, but easier, but that was my best
Hello, I am following you from Iraq. Can you explain this question? It came in the monthly exam at our school.
Prove z^2-1/z^2+1=itanø , z = cosø + isinø
This is wrong statement bcz put phi equals 0
Then LHS is 1 but RHS is 0
n^4 + 2n^2 + 1 = (n^2+1)^2 < n^4 + 3n^2 + 2 < n^4 + 4n^2 + 4 = (n^2 + 2)^2 therefore, we will need a perfect square that is between two consecutive perfect squares.
I love seeing another person who has a passion for math and teaching and is also religious.
n^2 and n^4 are always equivalent mod 4 (0^2=0^4=0, 1^2=1^4=1, 2^2=2^4=0, 3^2=3^4=1).
Thus the expression is equivalent to n^2+3n^2+2 = 4n^2+2 = 2 mod 4.
This also proves the evenness of the expression since any integer that is equivalent to 2 mod 4 is even.
If the product of 2 consecutive numbers is a square, since consecutive numbers are coprime, both numbers must be squares. But the only consecutive squares are 0 and 1. Even without being able to factor it, if k^2 = n^4 + 3n^2 + 2, 4k^2 = 4n^4 + 12n^2 + 8, (2n^2 + 3)^2 - (2k)^2 = 1.
The first thing that came to my mind is that the expression is also equal to (n^2+1)^2 + (n^2+1), ie it’s a perfect square plus the base of that perfect square , in the form m^2 + m
But the next perfect square after m^2 is 2m+1 apart. There is no way m^2+m is a perfect square
Let us prove that there does not exist an m in Z satisfying the equation m^2 = n^4 + 3n^2 + 2. To do this, we will examine the equation modulo 3. If no solution m exists in Z/3Z, then no solution can exist in Z either. While the converse is not necessarily true, this approach suffices to establish our claim.
First, recall that for the prime number 3, Wilson's theorem implies a^2 ≡ 1 mod 3. Using this, we reduce the given equation modulo 3. Since 3n^2 ≡ 0 mod 3, the equation becomes:
m^2 ≡ n^4 + 2 mod 3.
Noting that m^2 ≡ 1 mod 3, we rearrange to obtain:
1 ≡ n^4 + 2 mod 3,
or equivalently,
n^4 ≡ -1 mod 3. {At this point, we also note that
Z/3Z is a finite field (a finite commutative ring with multiplicative inverses for all non-zero elements). Therefore, all arithmetic operations are well-defined and the equation we are considering is valid in this context.}
Now, let us determine whether -1 is a quadratic residue modulo 3. Since 3 is a prime of the form 4k + 3, it follows that -1 is a quadratic non-residue modulo 3. This means there is no n in Z/3Z such that n^2 ≡ -1 mod 3. Consequently, it is impossible for n^4 ≡ -1 mod 3 to hold.
Thus, we reach a contradiction: the equation m^2 = n^4 + 3n^2 + 2 has no solutions in Z/3Z, and therefore, it has no solutions in Z. This completes the proof.
Before viewing the video: n⁴ + 3n² + 2 ≡ 2 (mod 4) for all n, but perfect square are either 0 or 1 (mod 4).
I replaced n² with m and to make m² + 3m + 2 = k², and then solved for m² + 3m + 2 - k² = 0 via the quadratic equation. For k² and m are perfect squares and must both be positive integers. But the Q.E. gave me m = (-3 +/- sqrt(k² + 1)) / 2, which is k² is a square, forces m to be irrational or negative.
I Have a question, Do you also use HAGOMORO Chalks!?
We could generalize the problem: for what values of k is n^4+3n^2+k a perfect square?
For all integer n, n² = 0 or 1 mod 4, thus n^4+3n²=n²(n²+3)=0 mod 4 thus n^4+3n²+2 = 2 mod 4 for all integer n. However, there is no square integer which is 2 modulo 4.
i think we can just end it on the first step
the expression we have boils down to (n^2 + 1)(n^2 + 2)
these are consecutive integers for all integral n, and the product of two consecutive integers can never be a perfect square, hence proven no?
14:30 can’t an odd number also be represented by (2k-1)? Does the proof still work by using that?
yes it does
(n^2+1)^2
Hi, since the comment section is the only way that outsiders can have a chance to contact you, would you please, consider putting a video about integrating probability functions? Currently I am having trouble integrating (f(x)/(1-F(x)))^0.5 over dx, where f(x)-pdf and F-cdf and the whole function is a conditional probability squared. Any useful hint would be appreciated!!
How did you reach the conclusion that began with 'since................'?
@ you did not put an email in the about section of your channel or in the video description, so I thought that only your patreons can directly contact you.
n^4+2n^2+n^2+2=(n^2+1)(n^2+2)🎉
Can u provide questions on ntg roots of unity
Reducing mod 4 is quick and easy
Thanks Sir 🙏
5:00 this way is kinda wrong, you forgot the absolute value on m (n²+1 < |m| < n²+2), or said that m is a non negative integer
it’s kinda implied that m is non-negative because it is found between n^2 + 1 and n^2 + 2, and n^2 can never be negative, meaning that m can never be negative either.
Factors (n^2+2)(n^2+1) so is a perfect square if and only if 2 = 1 a contradiction.
m must be between n^2+1 and n^2+2. n^2+1
Perfect square or not, its a perfect equation with four distinct answers. n=i,-i,2^1/2i and -2^1/2i.
sir. 3rd way to integrate 1/(1+x^4)
Coming soon
(n² +1)(n² +1 +1) = N (N +1)
Product of two side by side numbers couldn't be a square!
Is my proof wrong, if so why? PLS HELP(im in grade 10)
n^4 + 3n^2 + 2
(quadratic in terms of n^2)
if it is a perfect square(assume),
D= b^2 - 4ac = 0 (as roots are equal and real)
D=9 - 4(1)(2)=0
D= 9 - 8 = 0
D = 1 =0 (contradiction)
so original statement is wrong and it ain't a square number
Sir please do a limit question which was came in
JEE Advanced 2014 shift-1 question number 57
it's a question of a limit
lim as x-->1
[{-ax + Sin(x-1) + a}/{x + Sin(x-1) - 1}]^[{1-x}/{1-√x}] = 1/4
You have to find the greatest value of a
It has 2 possible answers 0 and 2
But I want the reason that why should I reject 2 and accept 0
Because final answer is 0
Please help 😢
I love numbers!
Claim: n^2(n^2+3) is divisible by 4 case 1: n is even (2k)^2((2k)^2+3)=4k^2((2k)^2+3) n is odd (2k+1)^2((2k+1)+3)=(2k+1)(4k^2+4k+4)=4(2k+1)^2(k^2+k+1)
Why aren't u using discriminant to prove taking n^2=t
For all integers,prove that n^4+3n^2+2 is never a perfect square. First method Suppose there exist some integer m such that m^2=n^4+3n^2+2=(n^2+1)(n^2+2) x^2=4*9 x=±6
X^2=16 then x is not between 4 and 4
Case 2: n is odd n^2 is odd n^2+3 is even n^2(n^2+3)+2 is even.
those who stop learning stop living
buh-bye
Claim: n^4+3n^2+2 is always even. n^4+3n^2+2=n^2(n^2+3)+2 Case 1:n is even n^2 is even n^2(n^2+3)+2 is even.
Since n^4+3n^2 is divisible by 4, n^4+3n^2+2 is never a multiple of 4 since 2
0:47 I miss the smile.....❤❤😅😅
n^4+3n^2+2 is always even.
Love it!!
Do you think this solution would work?
Suppose that n^4 + 3n^2 + 2 = k^2 for some k.
Now we can show that because the factorisation of n^4 + 3n^2 + 2 is (n^2 + 1)(n^2 + 2), the expression minimally has a value of 2 at n = 0. [Or you can differentiate the expression, whichever feels more rigorous].
2 >1^2 and 0^2, so this expression can never equal those squares.
Now we consider 2 cases of k^2, where it is a prime square [so k is prime like 2,3,5,7,11], and k is a composite square, [so k is composite like 4,6,8,9,10]
If k is a prime square, it's only factors are 1,k and k^2, so we could try to see if any values of n could make a value of k exist.
Case 1: n^2 + 1 = 1, n^2 + 2 = k^2
Eq 1 shows that n = 0, so we'd get k^2 = 2 ,but k wouldn't be an integer as 2 isn't a perfect square. So this case is out.
Case 2: n^2 + 1 = k, n^2 + 2 = k
It is "trivial", but trying to solve the two equations yields 1 = 2, which means these 2 expressions can never equal each other at the same value of n.
So the expression can never be a prime square. No prime values of k exist, so no n exists for which the expression is a perfect square.
For non prime squares, its factors are 1,k and k^2, alongside any other factors between 1 and k, and k and k^2.
We assume that the two factors n^2 + 2 and n^2 + 1 are factors that pair up to make a square that are not equal to each other [as we shown earlier]. So this means that:
n^2 + 1 < k < n^2 + 2, where k is the non-prime factor of the non-prime square.
This makes sense because factors will always pair up in a number. Since squares have 1 double counted factor, any factor pairs must lie on opposite sides of k.
We run into another issue however. n^2 + 1 - (n^2 + 1) = 1, so these numbers are just 1 apart. There is no possible way to sandwich a value of k for any n, because these 2 are consecutive integers. So no value of non-prime k exists as well, so no n exists for which the expression is a perfect square.
Since we showed that no value of k can exist for primes and non primes, as well as 0,1 this covers all the positive integers, and since k^2 = (-k^2), this covers the entire number range as well, so no value of k exists for which an integer n can produce a perfect square. As completion, Because the function is even, this also doubles as covering all integers, as f(n) = f(=n), so since we proved it for all positive n, any negative n will have the same result.
Hence, no value of n can produce a perfect square. QED.
Therefore n^4+3n^2+2 is not a perfect square for all integers.
f( n ) = n^4 + 3n^2 + 2 =
= ( n^2 + 1 )( n^2 + 2 )
n € Z
( n^2 + 1 )^2 < f( n ) < ( n^2 + 2 )^2
k = n^2 + 1 € IN
k^2 < f( n ) < ( k + 1 )^2
f( n ) no € IN^2 , n € Z
❤❤❤❤
Sqrt[20736]=144 perfect square root example.
n^4+3n^2+2=(n^2+1)(n^2+2) It’s in my head.
n^4+3n^2 is always divisible by 4.
Sqrt[65536]=256 another perfect square root.
Sir,please solve this problem. The broblem is taken from Bdmo.
𝒇(𝒏) = 𝟏𝟎^𝒏 − (𝟓 + √𝟏𝟕)^n
−(𝟓 − √𝟏𝟕)^n
is a function which is valid for all integers 𝒏 greater
than 𝟏. Prove that, 𝒇(𝒏) is always perfectly divisible by 𝟐^n+1.
I love the first proof. So simple.👍🏾💙🤍❤️