2:25 Many people including me, would ask the question about how to realize splitting RHS as 2 & 9 and then 3 & 4. Well this is the simplest way to understand that part - Just adding and subtracting the two equations and dividing result by 2 gives 9 and 4 respectively. Since square root operator is involved in equations, taking square root of resulting values gives 3 and 2 respectively. Again the question goes like - Why TAKE SQUARE ROOT??? For all such and further questions, you may refer exceptional minds like Ramanujan, as this seems out of bounds for people like me. However, I gave little thought and attempted to understand and share about the splitting part at 2:25. So you may appreciate if possible. Thank you and love you all. 😘🤗😊
You don't have to be a Ramanujan to understand what is being done here or to come up with this yourself. In fact, this is all basic and elementary algebra, _but_ the way in which this is presented in the video as well as the motivations for doing this are not made very clear. Let us start by reviewing some elementary properties of polynomials. If P(x) is a polynomial in x of degree n > 0 then all of the terms of this polynomial are of the form cᵢxⁱ where cᵢ are constants for i = 0..n. If we substitute x = a in this polynomial P(x) for any a then P(a) will likewise be a sum of terms which are all of the form cᵢaⁱ for i = 0..n. But if we then subtract P(a) from P(x) the difference P(x) − P(a) will be a sum of terms of the form cᵢxⁱ − cᵢaⁱ = cᵢ(xⁱ − aⁱ) for i = 1..n because there no longer is a constant term since c₀x⁰ − c₀a⁰ = c₀·1 − c₀·1 = c₀ − c₀ = 0. For any positive integer i, (xⁱ − aⁱ) contains a factor (x − a) since we have (x¹ − a¹) = (x − a)·1 (x² − a²) = (x − a)(x + a) (x³ − a³) = (x − a)(x² + ax + a²) (x⁴ − a⁴) = (x − a)(x³ + ax² + a²x + a³) ... and so on. It is easy to prove by induction that (xⁱ − aⁱ) contains a factor (x − a) for any positive integer i. If (xⁱ − aⁱ) contains a factor (x − a) for any given i then (xⁱ⁺¹ − aⁱ⁺¹) = xⁱx − xⁱa + xⁱa − aⁱa = xⁱ(x − a) + a(xⁱ − aⁱ) is the sum of two terms which each contain a factor (x − a) meaning that the (xⁱ⁺¹ − aⁱ⁺¹) itself also contains a factor (x − a). Alternatively, you can prove this directly by multiplying (xⁱ⁻¹ + axⁱ⁻² + ... + aⁱ⁻²x + aⁱ⁻¹) by (x − a) and noting that all products cancel except xⁱ⁻¹x − aⁱ⁻¹a = xⁱ − aⁱ. Clearly, since all terms of the difference P(x) − P(a) contain a factor (x − a) this means that P(x) − P(a) itself contains a factor (x − a), that is, there exists a polynomial Q(x) such that P(x) − P(a) = (x − a)·Q(x) and therefore P(x) = (x − a)·Q(x) + P(a) If, for any x ≠ a, we divide both sides by (x − a) this can be written as P(x)/(x − a) = Q(x) + P(a)/(x − a) meaning that the remainder of the division of a polynomial P(x) by (x − a) is simply P(a) and the quotient is some polynomial Q(x). This is the famous _Polynomial remainder theorem_ The remainder of the division of a polynomial P(x) by a linear polynomial x − a is equal to P(a). In the special case where x = a happens to be a zero of the polynomial P(x), that is, if P(a) = 0, then P(x) = (x − a)·Q(x) + P(a) reduces to P(x) = (x − a)·Q(x) That is, if x = a is a zero of a polynomial P(x), then P(x) contains a factor (x − a). It is trivial to prove that the converse is also true: if P(x) contains a factor (x − a), that is, if P(x) = (x − a)·Q(x) for some polynomial Q(x), then P(a) = (a − a)·Q(a) = 0·Q(a) = 0 so x = a is a zero of P(x). This is the famous _Factor theorem_ A polynomial P(x) of a degree greater than zero contains a factor (x − a) if and only if x = a is a zero of P(x). Before we move on to the problem in the video let us do a simple application of this theorem. Suppose we want to solve the cubic equation x³ − 2x² + 1 = 0 Now, we could use the cubic formula, but generally it is a beter idea to start by looking for rational solutions, because if we can find a rational solution x = r, then the factor theorem tells us we can factor out (x − r) at the left hand side and the other factor will then be a quadratic polynomial. The rational root theorem (which I will not discuss and prove here) tells us that any _potential_ rational solutions of this equation must be integers and must be divisors of the constant term 1, so we only need to test x = 1 and x = −1. And, sure enough, x = −1 is not a solution since (−1)³ − 2·(−1)² + 1 = −2, but x = 1 _is_ a solution since 1³ − 2·1² + 1 = 0 This means that according to the factor theorem (x − 1) is a factor of x³ − 2x² + 1. To find the other factor Q(x), that is, the quotient of x³ − 2x² + 1 and (x − 1), there are several methods. We can do a long division, a short division or a synthetic division or use Horners scheme but these methods can be rather cumbersome. But there is is very simple method to factor out (x − 1) inspired by the proof of the polynomial remainder theorem we have discussed. If we subtract 1³ − 2·1² + 1 = 0 from the equation x³ − 2·x² + 1 = 0 we get (x³ − 1³) − 2·(x² − 1²) = 0 So now we have a difference of two cubes (x³ − 1³) which contains a factor (x − 1) since x³ − 1³ = (x − 1)(x² + x + 1), a difference of two squares (x² − 1²) which contains a factor (x − 1) since x² − 1² = (x − 1)(x + 1) and no constant term on the left hand side. Replacing (x³ − 1³) with (x − 1)(x² + x + 1) and (x² − 1²) with (x − 1)(x + 1) we get (x − 1)(x² + x + 1) − 2·(x − 1)(x + 1) = 0 and taking out the common factor (x − 1) we have (x − 1)((x² + x + 1) − 2·(x + 1)) = 0 which simplifies to (x − 1)(x² − x − 1) = 0 Now, a product of any number of factors can be zero if and only if (at least) one of the factors is itself zero, so for any solution of our cubic equation other than x = 1 we must have x² − x − 1 = 0 That is, any solutions of the cubic equation x³ − 2x² + 1 = 0 other than x = 1 are solutions of this quadratic equation. The converse is also true, any solution of the quadratic equation x² − x − 1 = 0 will be a solution of the cubic equation x³ − 2x² + 1 = 0 since x³ − 2x² + 1 = (x − 1)(x² − x − 1) is zero if x² − x − 1 is zero. We can also tell, without solving the quadratic equation, that x² − x − 1 cannot have any rational zeros since x = 1 is the only rational zero of x³ − 2x² + 1 = (x − 1)(x² − x − 1). We can solve the quadratic equation x² − x − 1 = 0 using completion of the square or the quadratic formula and then we find that x = ½ + ½√5 and x = ½ − ½√5 are the solutions of the quadratic equation x² − x − 1 = 0 and these are therefore also solutions of the cubic equation x³ − 2x² + 1 = 0.
Now let us examine the problem in the video. We are asked to solve the system √x + y = 11 √y + x = 7 where x and y are assumed to be non-negative real numbers. We can start by looking for integer solutions. First of all, it is clear that neither x nor y can be zero. If we put x = 0 then the first equation implies y = 11 whereas the second equation then reduces to √y = 7 which implies y = 49, which is a contradiction. Similarly, if we put y = 0 then the first equation reduces to √x = 11 which implies x = 121 whereas the second equation then implies x = 7 which is again a contradiction. Now, since from the first equation we have √x = 11 − y and from the second equation we have √y = 7 − x it is clear that for any solution in positive integers both x and y must be perfect squares. Moreover, x must be smaller than 7 and y must be smaller than 11. This means that, for any potential integer solution, x can only be 1 or 4 and y can only be 1 or 4 or 9. It is now easy to find that x = 4 and y = 9 is indeed the only integer solution. But that was the easy part. We should now examine if the system has any other solutions not in integers. If it does, we should _find_ all other solutions, and if there are no other solutions we should _prove_ that there are none. But how can we do that? This problem differs from a polynomial equation since we now have a system of two equations in two variables x and y, not just a single equation in a single variable. It also differs since the equations themselves are not polynomials in the variables x and y since they contain square roots with these variables. But we can change this system into a system of polynomials by substituting √x = u and √y = v since that will get rid of the square roots. If we do this, the system becomes u + v² = 11 v + u² = 7 Since u = √x and v = √y and since x = 4, y = 9 is a solution of the original system, it follows that u = 2, v = 3 is a solution of this system in u and v, and indeed we have 2 + 3² = 11 3 + 2² = 7 In order to find other solutions of this system, the idea is to factor out (u − 2) and (v − 3) just like we factored out (x − 1) with our cubic equation, since the remaining factor(s) should then provide the condition(s) which any other solutions must satisfy. Now, if we subtract 2 + 3² = 11 from u + v² = 11 and subtract 3 + 2² = 7 from v + u² = 7 like we did with our cubic equation, we get (u − 2) + (v² − 3²) = 0 (v − 3) + (u² − 2²) = 0 Now we have a difference of two squares (v² − 3²) = (v − 3)(v + 3) and a difference of two squares (u² − 2²) = (u − 2)(u + 2) so we can rewrite the system as (u − 2) + (v − 3)(v + 3) = 0 (v − 3) + (u − 2)(u + 2) = 0 Now let us see what we have. The right hand sides of both our equations are zero, and we have a common factor (u − 2) and a common factor (v − 3) _but_ these are each in different equations, not in different terms of the same equation, so we cannot factor these out. How do we get around this? Well, the idea is to create a _single_ equation from these two equations which contains only terms with both (u − 2) and (v − 3) so we can factor both out. We cannot just multiply both equations as they stand because then we will also get a term which contains (u − 2)² but not (v − 3) and a term which contains (v − 3)² but not (u − 2) and then we still cannot factor out (u − 2)(v − 3). It is fine if the term (u − 2) gets multiplied by the term (v − 3) and it is fine if the term (v − 3)(v + 3) gets multiplied by (u − 2)(u + 2) but we do not want to multiply (u − 2) by (u − 2)(u + 2) or (v − 3) by (v − 3)(v + 3). So, what we need to do first is make shure that the terms (u − 2)(u + 2) and (v − 3)(v + 3) are on one side while the terms (u − 2) and (v − 3) are on the other side. To do that, we can bring the terms (u − 2) and (v − 3) over to the right hand side of the equations, which gives (v − 3)(v + 3) = −(u − 2) (u − 2)(u + 2) = −(v − 3) Now, when we multiply both equations we get (u − 2)(v − 3)(u + 2)(v + 3) = (u − 2)(v − 3) which is a condition which any solution of our system in u en v must satisfy. We now have two terms which contain a common factor (u − 2)(v − 3) but they are on opposite sides of this equation. If we want to take the common factor out we must first get both terms on the same side which we can do by subtracting (u − 2)(v − 3) from both sides which gives (u − 2)(v − 3)(u + 2)(v + 3) − (u − 2)(v − 3) = 0 and taking out the common factor (u − 2)(v − 3) then gives (u − 2)(v − 3)((u + 2)(v + 3) − 1) = 0 so we find that any solution of the system in u and v must satisfy (u − 2)(v − 3) = 0 or (u + 2)(v + 3) − 1 = 0 Now clearly (u − 2)(v − 3) can only be zero if at least one of its two factors is zero, that is, if either u = 2 or v = 3 is true or if both are true. But we already know that u = 2 implies v = 3 and that v = 3 implies u = 2, so for any _other_ potential solutions of our system in u and v we must have both u ≠ 2 and v ≠ 3 which implies (u − 2)(v − 3) ≠ 0. Therefore, any solution of the system in u and v _other_ than u = 2, v = 3 must satisfy (u + 2)(v + 3) − 1 = 0, that is (u + 2)(v + 3) = 1 However, it is clear that there can exist no positive real numbers u and satisfying this condition, because for any positive real numbers u and v we have u + 2 > 2 and v + 3 > 3 and therefore (u + 2)(v + 3) > 6 So, we have proved that no positive real solutions of the system u + v² = 11, v + u² = 7 other than u = 2, v = 3 can exist and, consequently, no positive real solutions other than x = 4, y = 9 of the system √x + y = 11, √y + x = 7 can exist. It will be clear now that to reach this conclusion we did not actually have to perform the substitutions u = √x and v = √y because we can rewrite the original system √x + y = 11 √y + x = 7 as √x + (√y)² = 11 √y + (√x)² = 7 and consider this as a system of polynomials in √x and √y. Using the exact same method we can then find that any solution of this system must satisfy (√x − 2)(√y − 3)((√x + 2)(√y + 3) − 1) = 0 Consequently, any solution other than x = 4, y = 9 would have to satisfy (√x + 2)(√y + 3) = 1 which is impossible for any non-negative real numbers x and y.
I assumed x and y to be integers, so i didn't really prove that there were no other real solutions. But I did not guess/brute force. I set sqrt(x)=a and sqrt(y)=b, eqns become: a + b^2 = 11 b + a^2 = 7 Subtract eqn 2 from eqn 1: a - b + b^2 - a^2 = 4, difference of squares a - b + (b-a)(b+a) = 4, collect like terms (a-b)(1-a-b) = 4. a and b are integers so you just do the listing factors method, note you also include negative integers that multiply to 4. Overall you check 6 cases total but you quickly reject all but 1 because a and b are positive integers. You're left with a=2, b=3 or x=4 and y=9. So in a universe where you cant brute force this, I guess this method is better + it finds _all_ integer solutions.
1:25 It doesn't work that way, but with the help of the rational root theorem we can at least find one solution x=4 giving us the pair (4,9) without guessing. Then of course, as we are not able to find the other 3 solutions (those are irrational), this approach is missing to show, that (4,9) is the only possible pair.
y = 11 - sqrt(x) y = (7-x)^2 one falls, one rises -> one one intersection and x=2^2 y=3^2 solution, being 2 small squares, is just too obvious not to check
As others have shown, you can just subtract eqn 2 from eqn 1 and factorize (y - x) to (√y - √x)(√y + √x). However this method does not prove that there are no other non-integer real soluntions (or not). I'll be interested to learn such a method.
Interesting how the author shows the vast array of techniques a mathematician must be accustomed to that when something is easily visible to the eye he'll avoid going to the usuals....ie in this video he avoids substituting variables ,something he commonly does OR he rarely uses summation symbol but he did yesterday
Very beautiful and ingenius which should be expected from Ramanujan!
2:25 Many people including me, would ask the question about how to realize splitting RHS as 2 & 9 and then 3 & 4.
Well this is the simplest way to understand that part -
Just adding and subtracting the two equations and dividing result by 2 gives 9 and 4 respectively.
Since square root operator is involved in equations, taking square root of resulting values gives 3 and 2 respectively.
Again the question goes like -
Why TAKE SQUARE ROOT???
For all such and further questions, you may refer exceptional minds like Ramanujan, as this seems out of bounds for people like me. However, I gave little thought and attempted to understand and share about the splitting part at 2:25. So you may appreciate if possible.
Thank you and love you all.
😘🤗😊
You don't have to be a Ramanujan to understand what is being done here or to come up with this yourself. In fact, this is all basic and elementary algebra, _but_ the way in which this is presented in the video as well as the motivations for doing this are not made very clear.
Let us start by reviewing some elementary properties of polynomials. If P(x) is a polynomial in x of degree n > 0 then all of the terms of this polynomial are of the form cᵢxⁱ where cᵢ are constants for i = 0..n. If we substitute x = a in this polynomial P(x) for any a then P(a) will likewise be a sum of terms which are all of the form cᵢaⁱ for i = 0..n. But if we then subtract P(a) from P(x) the difference P(x) − P(a) will be a sum of terms of the form cᵢxⁱ − cᵢaⁱ = cᵢ(xⁱ − aⁱ) for i = 1..n because there no longer is a constant term since c₀x⁰ − c₀a⁰ = c₀·1 − c₀·1 = c₀ − c₀ = 0.
For any positive integer i, (xⁱ − aⁱ) contains a factor (x − a) since we have
(x¹ − a¹) = (x − a)·1
(x² − a²) = (x − a)(x + a)
(x³ − a³) = (x − a)(x² + ax + a²)
(x⁴ − a⁴) = (x − a)(x³ + ax² + a²x + a³)
...
and so on. It is easy to prove by induction that (xⁱ − aⁱ) contains a factor (x − a) for any positive integer i. If (xⁱ − aⁱ) contains a factor (x − a) for any given i then (xⁱ⁺¹ − aⁱ⁺¹) = xⁱx − xⁱa + xⁱa − aⁱa = xⁱ(x − a) + a(xⁱ − aⁱ) is the sum of two terms which each contain a factor (x − a) meaning that the (xⁱ⁺¹ − aⁱ⁺¹) itself also contains a factor (x − a). Alternatively, you can prove this directly by multiplying (xⁱ⁻¹ + axⁱ⁻² + ... + aⁱ⁻²x + aⁱ⁻¹) by (x − a) and noting that all products cancel except xⁱ⁻¹x − aⁱ⁻¹a = xⁱ − aⁱ.
Clearly, since all terms of the difference P(x) − P(a) contain a factor (x − a) this means that P(x) − P(a) itself contains a factor (x − a), that is, there exists a polynomial Q(x) such that
P(x) − P(a) = (x − a)·Q(x)
and therefore
P(x) = (x − a)·Q(x) + P(a)
If, for any x ≠ a, we divide both sides by (x − a) this can be written as P(x)/(x − a) = Q(x) + P(a)/(x − a) meaning that the remainder of the division of a polynomial P(x) by (x − a) is simply P(a) and the quotient is some polynomial Q(x). This is the famous
_Polynomial remainder theorem_
The remainder of the division of a polynomial P(x) by a linear polynomial x − a is equal to P(a).
In the special case where x = a happens to be a zero of the polynomial P(x), that is, if P(a) = 0, then P(x) = (x − a)·Q(x) + P(a) reduces to
P(x) = (x − a)·Q(x)
That is, if x = a is a zero of a polynomial P(x), then P(x) contains a factor (x − a). It is trivial to prove that the converse is also true: if P(x) contains a factor (x − a), that is, if P(x) = (x − a)·Q(x) for some polynomial Q(x), then P(a) = (a − a)·Q(a) = 0·Q(a) = 0 so x = a is a zero of P(x). This is the famous
_Factor theorem_
A polynomial P(x) of a degree greater than zero contains a factor (x − a) if and only if x = a is a zero of P(x).
Before we move on to the problem in the video let us do a simple application of this theorem. Suppose we want to solve the cubic equation
x³ − 2x² + 1 = 0
Now, we could use the cubic formula, but generally it is a beter idea to start by looking for rational solutions, because if we can find a rational solution x = r, then the factor theorem tells us we can factor out (x − r) at the left hand side and the other factor will then be a quadratic polynomial. The rational root theorem (which I will not discuss and prove here) tells us that any _potential_ rational solutions of this equation must be integers and must be divisors of the constant term 1, so we only need to test x = 1 and x = −1. And, sure enough, x = −1 is not a solution since (−1)³ − 2·(−1)² + 1 = −2, but x = 1 _is_ a solution since
1³ − 2·1² + 1 = 0
This means that according to the factor theorem (x − 1) is a factor of x³ − 2x² + 1. To find the other factor Q(x), that is, the quotient of x³ − 2x² + 1 and (x − 1), there are several methods. We can do a long division, a short division or a synthetic division or use Horners scheme but these methods can be rather cumbersome. But there is is very simple method to factor out (x − 1) inspired by the proof of the polynomial remainder theorem we have discussed. If we subtract 1³ − 2·1² + 1 = 0 from the equation x³ − 2·x² + 1 = 0 we get
(x³ − 1³) − 2·(x² − 1²) = 0
So now we have a difference of two cubes (x³ − 1³) which contains a factor (x − 1) since x³ − 1³ = (x − 1)(x² + x + 1), a difference of two squares (x² − 1²) which contains a factor (x − 1) since x² − 1² = (x − 1)(x + 1) and no constant term on the left hand side. Replacing (x³ − 1³) with (x − 1)(x² + x + 1) and (x² − 1²) with (x − 1)(x + 1) we get
(x − 1)(x² + x + 1) − 2·(x − 1)(x + 1) = 0
and taking out the common factor (x − 1) we have
(x − 1)((x² + x + 1) − 2·(x + 1)) = 0
which simplifies to
(x − 1)(x² − x − 1) = 0
Now, a product of any number of factors can be zero if and only if (at least) one of the factors is itself zero, so for any solution of our cubic equation other than x = 1 we must have
x² − x − 1 = 0
That is, any solutions of the cubic equation x³ − 2x² + 1 = 0 other than x = 1 are solutions of this quadratic equation. The converse is also true, any solution of the quadratic equation x² − x − 1 = 0 will be a solution of the cubic equation x³ − 2x² + 1 = 0 since x³ − 2x² + 1 = (x − 1)(x² − x − 1) is zero if x² − x − 1 is zero. We can also tell, without solving the quadratic equation, that x² − x − 1 cannot have any rational zeros since x = 1 is the only rational zero of x³ − 2x² + 1 = (x − 1)(x² − x − 1). We can solve the quadratic equation x² − x − 1 = 0 using completion of the square or the quadratic formula and then we find that x = ½ + ½√5 and x = ½ − ½√5 are the solutions of the quadratic equation x² − x − 1 = 0 and these are therefore also solutions of the cubic equation x³ − 2x² + 1 = 0.
Now let us examine the problem in the video. We are asked to solve the system
√x + y = 11
√y + x = 7
where x and y are assumed to be non-negative real numbers. We can start by looking for integer solutions. First of all, it is clear that neither x nor y can be zero. If we put x = 0 then the first equation implies y = 11 whereas the second equation then reduces to √y = 7 which implies y = 49, which is a contradiction. Similarly, if we put y = 0 then the first equation reduces to √x = 11 which implies x = 121 whereas the second equation then implies x = 7 which is again a contradiction.
Now, since from the first equation we have √x = 11 − y and from the second equation we have √y = 7 − x it is clear that for any solution in positive integers both x and y must be perfect squares. Moreover, x must be smaller than 7 and y must be smaller than 11. This means that, for any potential integer solution, x can only be 1 or 4 and y can only be 1 or 4 or 9. It is now easy to find that x = 4 and y = 9 is indeed the only integer solution.
But that was the easy part. We should now examine if the system has any other solutions not in integers. If it does, we should _find_ all other solutions, and if there are no other solutions we should _prove_ that there are none. But how can we do that?
This problem differs from a polynomial equation since we now have a system of two equations in two variables x and y, not just a single equation in a single variable. It also differs since the equations themselves are not polynomials in the variables x and y since they contain square roots with these variables. But we can change this system into a system of polynomials by substituting √x = u and √y = v since that will get rid of the square roots. If we do this, the system becomes
u + v² = 11
v + u² = 7
Since u = √x and v = √y and since x = 4, y = 9 is a solution of the original system, it follows that u = 2, v = 3 is a solution of this system in u and v, and indeed we have
2 + 3² = 11
3 + 2² = 7
In order to find other solutions of this system, the idea is to factor out (u − 2) and (v − 3) just like we factored out (x − 1) with our cubic equation, since the remaining factor(s) should then provide the condition(s) which any other solutions must satisfy. Now, if we subtract 2 + 3² = 11 from u + v² = 11 and subtract 3 + 2² = 7 from v + u² = 7 like we did with our cubic equation, we get
(u − 2) + (v² − 3²) = 0
(v − 3) + (u² − 2²) = 0
Now we have a difference of two squares (v² − 3²) = (v − 3)(v + 3) and a difference of two squares (u² − 2²) = (u − 2)(u + 2) so we can rewrite the system as
(u − 2) + (v − 3)(v + 3) = 0
(v − 3) + (u − 2)(u + 2) = 0
Now let us see what we have. The right hand sides of both our equations are zero, and we have a common factor (u − 2) and a common factor (v − 3) _but_ these are each in different equations, not in different terms of the same equation, so we cannot factor these out. How do we get around this? Well, the idea is to create a _single_ equation from these two equations which contains only terms with both (u − 2) and (v − 3) so we can factor both out.
We cannot just multiply both equations as they stand because then we will also get a term which contains (u − 2)² but not (v − 3) and a term which contains (v − 3)² but not (u − 2) and then we still cannot factor out (u − 2)(v − 3). It is fine if the term (u − 2) gets multiplied by the term (v − 3) and it is fine if the term (v − 3)(v + 3) gets multiplied by (u − 2)(u + 2) but we do not want to multiply (u − 2) by (u − 2)(u + 2) or (v − 3) by (v − 3)(v + 3). So, what we need to do first is make shure that the terms (u − 2)(u + 2) and (v − 3)(v + 3) are on one side while the terms (u − 2) and (v − 3) are on the other side. To do that, we can bring the terms (u − 2) and (v − 3) over to the right hand side of the equations, which gives
(v − 3)(v + 3) = −(u − 2)
(u − 2)(u + 2) = −(v − 3)
Now, when we multiply both equations we get
(u − 2)(v − 3)(u + 2)(v + 3) = (u − 2)(v − 3)
which is a condition which any solution of our system in u en v must satisfy. We now have two terms which contain a common factor (u − 2)(v − 3) but they are on opposite sides of this equation. If we want to take the common factor out we must first get both terms on the same side which we can do by subtracting (u − 2)(v − 3) from both sides which gives
(u − 2)(v − 3)(u + 2)(v + 3) − (u − 2)(v − 3) = 0
and taking out the common factor (u − 2)(v − 3) then gives
(u − 2)(v − 3)((u + 2)(v + 3) − 1) = 0
so we find that any solution of the system in u and v must satisfy
(u − 2)(v − 3) = 0 or (u + 2)(v + 3) − 1 = 0
Now clearly (u − 2)(v − 3) can only be zero if at least one of its two factors is zero, that is, if either u = 2 or v = 3 is true or if both are true. But we already know that u = 2 implies v = 3 and that v = 3 implies u = 2, so for any _other_ potential solutions of our system in u and v we must have both u ≠ 2 and v ≠ 3 which implies (u − 2)(v − 3) ≠ 0. Therefore, any solution of the system in u and v _other_ than u = 2, v = 3 must satisfy (u + 2)(v + 3) − 1 = 0, that is
(u + 2)(v + 3) = 1
However, it is clear that there can exist no positive real numbers u and satisfying this condition, because for any positive real numbers u and v we have u + 2 > 2 and v + 3 > 3 and therefore
(u + 2)(v + 3) > 6
So, we have proved that no positive real solutions of the system u + v² = 11, v + u² = 7 other than u = 2, v = 3 can exist and, consequently, no positive real solutions other than x = 4, y = 9 of the system √x + y = 11, √y + x = 7 can exist.
It will be clear now that to reach this conclusion we did not actually have to perform the substitutions u = √x and v = √y because we can rewrite the original system
√x + y = 11
√y + x = 7
as
√x + (√y)² = 11
√y + (√x)² = 7
and consider this as a system of polynomials in √x and √y. Using the exact same method we can then find that any solution of this system must satisfy
(√x − 2)(√y − 3)((√x + 2)(√y + 3) − 1) = 0
Consequently, any solution other than x = 4, y = 9 would have to satisfy
(√x + 2)(√y + 3) = 1
which is impossible for any non-negative real numbers x and y.
nice one again
Show!!!👋👋👋👋
I assumed x and y to be integers, so i didn't really prove that there were no other real solutions. But I did not guess/brute force. I set sqrt(x)=a and sqrt(y)=b, eqns become:
a + b^2 = 11
b + a^2 = 7
Subtract eqn 2 from eqn 1:
a - b + b^2 - a^2 = 4, difference of squares
a - b + (b-a)(b+a) = 4, collect like terms
(a-b)(1-a-b) = 4. a and b are integers so you just do the listing factors method, note you also include negative integers that multiply to 4.
Overall you check 6 cases total but you quickly reject all but 1 because a and b are positive integers.
You're left with a=2, b=3 or x=4 and y=9.
So in a universe where you cant brute force this, I guess this method is better + it finds _all_ integer solutions.
1:25 It doesn't work that way, but with the help of the rational root theorem we can at least find one solution x=4 giving us the pair (4,9) without guessing. Then of course, as we are not able to find the other 3 solutions (those are irrational), this approach is missing to show, that (4,9) is the only possible pair.
y = 11 - sqrt(x)
y = (7-x)^2
one falls, one rises -> one one intersection
and x=2^2 y=3^2 solution, being 2 small squares, is just too obvious not to check
We can subtract 1 and 2 and factorize
x = 4, y = 9
Solved before looking at the video
As others have shown, you can just subtract eqn 2 from eqn 1 and factorize (y - x) to (√y - √x)(√y + √x). However this method does not prove that there are no other non-integer real soluntions (or not). I'll be interested to learn such a method.
Learnt yet another interesting solution.thanks.
In before all those genius people coming in and saying what an easy question this is... Please don't.
Hahaha....all those genius minds be guessing the solution.
This is genuinely tough tbh. I am not a genius though
if you want easy question go to the which is bigger sqrt(2) or cube root of 3 video.
Yes.
👍
Interesting how the author shows the vast array of techniques a mathematician must be accustomed to that when something is easily visible to the eye he'll avoid going to the usuals....ie in this video he avoids substituting variables ,something he commonly does OR he rarely uses summation symbol but he did yesterday
(4,9)
X=4, Y=9
(X,Y)=(4,9)
Y=9 x=4
I can solve it easily bro by imagining 4 and 9 as xand y
X=4,y=9
X= 4 y=9
sqrt(x)+sqrt(y)+x+y=18, guess 12+6 correct
Una soluzione é 4,9,un altra é 9,8 e 7,87....poi forse ce ne sono altre reali
(9.8..., 7.87...) isn't a solution to sqrt(y)+x=7. It's a false solution introduced by squaring one of the equations.
BRO CONTATTAMI