How real men solves a simple equation (when Ramanujan gets bored)

a problem from Ramanujan
radical equation
pair of equations
algebraic equations
#ramanujan
#euler
#olympiad

[anatakos]

The fact that we look for integer values should be stated at the beginning and not inferred at some point during the exercise

[celestiaaaa]

As a fellow guy who likes math, I solved this in 7 hours.

*It was so easy.*

[Dpak]

I solved this in 5 seconds . By forwarding to 7.00.

[stephenlpitts]

This is the most complicated over-the-top brute force solution that I could imagine. It’s much so much simpler: we can assume x and y are positive integers to see if we can find a quick solution since it is so easy to check all the possibilities in two minutes (proving its the only solution we can address afterwards). If they are not, then we have to resort to other methods. So both x and y can only be 1, 4 or 9 (any higher and it doesn’t work). And so their square roots can only be 1, 2 or 3. So we need a combination of x and y and their respective square roots which solves the two equations. One can quickly figure out that x=9 and y=4 does solve it and is the only combination here that does. In retrospect, the problem is wonderfully simple. In response to the Lazy Guy, good point, but honestly if you are a kid and now an adult, then please remember don’t lose your most important gift. Imagination. “The hard way” here is devoid of almost any of it.

[Skank_and_Gutterboy]

For this problem, I assumed that x and y are real numbers. Then I rearranged the first equation to: sqrt(x) = 7 - y, then squared both sides. Then x = y^2 - 14y + 49. For the 2nd equation, I rearranged it to: x = 11 - sqrt(y). Then I subtracted the rearranged 2nd equation from the first (so that you get an equation only in terms of y), rearranged that so that -sqrt(y) is on one side, then squared both sides. It yields the 4th order polynomial: y^4 - 28y^3 + 272y^2 - 1065y + 1444 = 0. The roots of this polynomial are then used to calculate x using the equation x = 11 - sqrt(y). Checking against the original equations, only (x, y) = (4, 9) satisfies the original equations. The other three roots of the 4th-order polynomial solved previously are extraneous.

[success]

First look: Both unknowns could be perfect square numbers. First guess 9 and 4. Solved. This is what I did in entrance exams(due to time constraint). But if you are kid then don't look for shortcuts. Learn the hard way first.

[vanijha]

We can also do that
Let x=a^2 and y=b^2
So equations will be
a+b^2=7(1)
a^2+b=11(2)
From(1)a=7-b^2then a^2=(7-b^2)^2
And from( 2) a^2=11-b
So
49+b^4-14b^2=11-b
or, b^4-14b^2-b+38
We can write this as
b^4-4b^2-10b^2+20b-19b+38
Now we can factor, after doing all process

b-2 is common, b=2
As taken y= b^2 y=4
2+x=11
x=9, y=4

[shivamonepiece]

Thank you bro very much no one was able to explain this better and easy than you

[programmerarnab]

as a fellow guy who do not understand math, . 🐐 i got nothing to say.. i know ramanuj and whatever you did on the video .. keep it up.. i don't understand it but those who do will appreciate it

[emilianobigozzi]

For integer solutions, both x and y must be perfect squares and from the first equation, y must be lower than 7.
The only possible values for y are 4 and 1 but the latter, fed into the second equation, yelds x equals to 10: not a square, so y=4 and x=9.

Besides, the system could have other solutions in the real numbers domain.

[creatorleo]

1. See the problem
2. Consider that x >=0 and y >=0, both would be integers to have an easy answer and also be squares
3. Start using 0, 1, 4, 9...
4. 9 and 4

[ravibhandari]

I am pretty sure that Ramanujan would not have solved it this way. One could straightaway see the answer as 4 and 9. Ultimately you are also doing trial and error, then why not do it from the beginning. The search space is further reduced by the fact that both x and y should be perfect squares (under the assunption that both x and y are integers).

[the-boy-who-lived]

If x and y are integers that means you just made a very simple problem too complex.
There is only few ways to get 7 with one integer (y in this case)
0+7, 1+6, 2+5, 3+4
In the next equation to get an integer (11 ) as a sum of an integer (x) and an another nunber (√y), That another number (√y) should be an integer as well. For √y to be an integer y should be a perfect square. Above I showed all the possible ways to get 7. If you look at it, you'll see only 0 and 4 are perfect squares. y must be equal to 0 or 4. But if y=0, √x=7 which means x=49. As 49+2 (x + √y) ≠ 11, x≠49, √x≠7, y≠0. Only option you are left with is y=4. If y=4 for √x+y to be equal to 7 √x has to be equal to 3. So x=3²=9
y=4, x=9

This explaination is long because it is so simplified, not because it is complex.

[trumanburbank]

You can use z = sqrt(x) + sqrt(y), w = sqrt(x) - sqrt(y), then wz = x - y. Subtracting one of the given equations by the other; x - y + sqrt(y) - sqrt(x) = 4 becomes wz - w = 4 becomes w(z - 1) = 4, etc.

[Laughing_Cat_Meme]

Edit: Turns out I was wrong; there is only one solution, the other 4 are extraneous due to squaring.


Original comment:


Actually there are total of 4 answers to this question;
all values of x satisfying the equation [0 = x⁴ - 44x³ + 712x² - 5107x + 12996] are the solutions too.
Because the Degree is 4, there are total of 4 solutions which are:
(9, 4)
(7.869..., 4.195...)
(12.848...., 3.415....)
(14.283..., 3.22...)

[DetCoAnimeFan]

As soon I saw this, first numbers that popped in my mind were 4 and 9. Possibly because I had played with squares a lot.

[jaconoorland]

The way of solving is need but I immediately saw that it is supposed to be 4 and 9. Sometimes the easy way of trial and error is better than the sophisticated way😂

[TheElevenBravo]

I remember this in 5th grade. That's when I told my mom I wanted to be home schooled

[arthur_paulino]

I solved it in 30 seconds with a naive assumption that x and y must be small quadratic numbers. My first attempt was to check if the equations matched for x=9 and y=4. Bingo.

[rajkumarlakhmani]

As soon as I saw Ramanujan, the scent of Number Theory surrounded me and bam!