South Africa Math Olympiad Question

Maths Olympiads are held all around the world to recognise students who excel in maths. The test is offered at many grade levels and provides them with numerous possibilities to win certifications, awards, and even scholarships for higher studies

[yamnitsky]

takes all of 3 seconds to solve by simple inspection: xy=35, the only factors of 35 that aren't trivial are 5 and 7. Checking 7^2-5^2 = 49-25=24. x=7, y=5. Similarly works with -7 and -5, so the answer is 12 or -12. No imaginary numbers necessary as far as I can tell.

[hnnh]

It's very simple
if xy=35
then y=35/×
and we can use it in
x^2-y^2=24

[ramitmazumder]

Substitution is an easier approach, get x in terms of y from eq 2, substitute in eq 1 and solve the quadratic equation

[pourianozari]

Alternatively, multiply the first eq by x^2
=> x^4 - x^2*y^2=24*x^2
=> x^4 - 24x^2 - (xy)^2=0
now plug xy=35 into this
=> (x^2)^2 - 24x^2 - 35^2 = 0
Note that this could be factored as follows: (Try looking at the factors in 35^2= 5*5*7*7 to find two numbers from this the difference of which would be 24 => You will find 5*5 and 7*7)
=> (x^2 - 49) (x^2+25) = 0 => x^2 = 49 or x^2=-25
You can take it from here that x can be either +7 or -7 which makes x+y +12 or -12.

[AlanCanon]

Solvable by inspection. 35 is nearly prime, it only has 5 and 7 as prime factors. So x, y = 35, 1 or 7, 5. x+y = 12. ps. forgot -7, -5 as solns.

[saragiotis]

When squaring both sides of an equation, it is possible to introduce solutions that do not satisfy the initial system equations. Although for this particular system of equations, this will not change anything, imho it would be better practice to check the final result with the initial system of equations

[kenkennio]

x+ y can be 4 different things: 12, -12, 2i and -2i

[mustafasabir]

I solved it in 2 seconds. First approach was to see 35 is prime so x and y could be 7 and 5. To make sure 7 sqaure - 5 square is also 24. So x + y is 12

[SerieusFrank]

it's just that 1x35 and 5x7 are the only quick integer opties to test out. I know x2-y2 is a symetric shape and xy=c a 1/x shape. thus 7, 5 and negatives are properly the real solutions.

[kanizfatema]

X^2 + Y^2= -70 is not valid, since the sum of the square of two number is always positive,
It must be X^2 + Y^2 = 70.

[MadChristoph]

Man, they made this simple math problem complex

[carlowood]

I'd subst. k=x+y, m=x-y, because that is the obvious thing to do. Then km = 24 and k^2 - m^2 = 4xy = 140. Multiply with k^2: k^4 - 140 k^2 - 24^2 = 0 = (k^2 - 144)(k^2 + 4) and we're done.

[yuki-dwjr]

If you define x + y ≡ k, you can consider that x and y are solutions of the following equation:
X^2 - (x+y) X + xy = X^2 - kX + 35 = 0.
So you can get x - y = √(k^2 - 140) by solving this equation.
You can obtain x^2 - y^2 = (x + y)(x - y) = k √(k^2 - 140) = 35, which means k = ±12, ±2i.

This problem is basic level for Japanese high school students.

[alfal]

Denote P = x^2, Q =- y^2
Then P and Q are the roots of t^2 – 24*t – 35^2 = 0

[ycnvqxn]

Set Y ²= a, the equation is converted to 1225/a-a=24→ 1225-a ²= 24a, a=49, so Y=7, X=5

[egghead]

very interesting. should take less than 2 sec to solve
xy=35
x^2-y^2 =24
Given that 35 is an odd number and a small number (a limited number sets to get 35), it is either 35*1 = 35 or 7*5 = 35. 7^2 - 5^2 = 24. So, x+y = 12.
Am I missing something ???

[d.m.]

Two more solutions.

X = 5i, Y = -7i
Or X = -5i, Y = 7i

Where i = square root of -1.

[omu_omuomu]

The question doesn't say X and Y are integers.. Therefore it's wrong to get X=7 and Y=5 immediately.

[danielrybuk]

Wow what a beautiful question, i got 7 and 5, but missed the complex solutions. Again beautiful question.

[joepollard]

I simply went to equation 2, and got 7x5 =35. Then 7²=49 and 5²=25, giving me 24. Then 7+5=12 Take the simple way. It can save you a bunch of time.