Algebra (1st math Thailand POSN 2014)

1) $\frac{169}{20} = 8.45$

2.1) Just expand

2.2) 12 integral solutions

3) $\displaystyle 0$

4) $2566$

Are these right?

For number 2, 2.1 can simply be proven by expanding but for 2.2 (somehow not using 2.1), x^2 - 34x + y^2 + 34y = 0, we can use completing the square to get (x - 17)^2 + (y + 17)^2 = 578 and express (x - 17)^2 = 578 - (y + 17)^2. Since x and y are positive, y > 0 and 289 < (y + 17)^2 <= 578. And this is true when 17 < y + 17 < 25. By checking one-by-one, the equation has integral solutions if y = 6 and gives x = 10 or 24. (10, 6) and (24, 6) are the only solutions.

Somehow the calculator said that the answer for 1) is

$\displaystyle 8.49 \leq x < 8.5$

For number 1, the answer is any real value of x: 7.58 <= x < 7.59

From this equation, there n of the terms in which the value is equal to x and 99 - n of the terms equivalent to x + 1. we bring up the equation: nx + (99 - n)(x + 1) = nx + 99x + 99 - nx - n = 761. And rewriting it into, 99x + 99 - n = 761 and 99x = 662 + n. Since x is integral, 99|(662 + n) and this is only possible when n = 31 and so.. x = 7. From here, we find out that 31 of terms in the sequence above has the floor of 7 and 68 of them has floor of 8. We compare x + 41/100 and x + 42/100 since x + 41/100 is the 31st term and x + 42/100 is the 32nd term where x + 41/100 < 8 <= x + 42/100. Rewriting the inequality gives 379/100 <= x < 759/100.

Trying 2.2)

$(x^{2}+y^{2})^{2} = 1156(x-y)^{2}$

$(x^{2}-y^{2})^{2} + 4x^{2}y^{2} = 1156(x-y)^{2}$

$(x-y)^{2}(x+y)^{2} + 4x^{2}y^{2} = 1156(x-y)^{2}$

$4x^{2}y^{2} = (1156-(x+y)^{2})(x-y)^{2}$

$2xy = \sqrt{(34^{2}-(x+y)^{2})}(x-y)$

Checking $2 < x+y < 34$ is going to be hardcore for me lol.

Calculator gives me $x+y = 16, 30$

Case1: $2xy = 30(x-y)$

$(x+15)(y-15) = -225$ which gives $(x,y) = (10,6)$.

Case2: $2xy = 16(x-y)$

$(x+8)(y-8) = -64$ which gives $(x,y) = (24,6)$.

In the 3 rd i got alpha + beta as 0 ? am i correct would anyone tell ?

For 2.1 a neater way would be to use complex numbers. $x^2+y^2 = (x +iy)(x-iy)$. Hence squaring it would give us $RHS = (x^2 - y^2 + i2xy)(x^2 - y^2 - i2xy)$. I believe what follows is obvious.

1. We have $\alpha^2 +1 = 3( \alpha + \beta)$ and $1 - 2(\alpha + \beta) = \alpha \beta$. Rewriting the second equation we have, $2 - 3( \alpha + \beta) = (1 - \alpha)(1 - \beta) \Rightarrow 2 - (\alpha^2 +1) = (1 - \alpha)(1 - \beta) \Rightarrow 1 + \alpha = 1 - \beta$. Thus the required value is zero.