$a,b$ are real numbers that satisfy
$$a^3 - 3ab^2 = 29$$
$$b^3 - 3a^2b = 34$$
What is the value of $a^2 + b^2$?
Meta puzzle: Which year did this problem appear?
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Solution:
The solution becomes simple if you think complex!
$$(a+ib)^3 = (a^3 - 3ab^2) - i(b^3 - 3a^2b)$$
Thus $$(a+ib)^3 = 29 - 34i$$
And so $$(a^2 + b^2)^3 = 29^2 + 34^2 = 1997$$
and
$$a^2 + b^2 = \sqrt[3]{1997}$$
Note that we basically have the identity
$$(a^2 + b^2)^3 = (a^3 - 3ab^2)^2 + (b^3 - 3a^2b)^2$$
A classic.
Show that integers of the form $2^m$ are the only positive integers that cannot be written as the sum of two or more consecutive positive integers.
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Solution:
Suppose we try to write $X$ as the sum of consecutive positive integers.
$$X = a + (a+1) + \dots + (a+n)$$
i.e.
$$2X = (n+2a)(n+1)$$
Note, since we need two or more, we must have that $n > 0$, and $n + 2a > n+1$.
Note that $n+2a$ and $n+1$ are of different parities.
If $X$ was a power of $2$, then the only way we can factor $2X$ into factors of different parities is if one of them is $1$. But we have $n+1 \ge 2$. So, powers of $2$ cannot be written in that form.
For any other $X = 2^m Y$ ($Y > 1$, odd), we can pick the factors to be $2^{m+1}$ and $Y$ and compute the appropriate $n$ and $a$.
The IOQM seems to have some nice problems, though the solutions given by the many of the Indian book authors/coaching classes are sometimes wrong and/or lacking for the harder problems.
Here is a problem where one of the books got the wrong answer, even though the problem isn't hard.
The altitude dropped on the hypotenuse of a right triangle is the integer $12$. Given that the hypotenuse and the perimeter of the triangle are integers too, what is the smallest possible length of the hypotenuse?
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Solution:
If $c$ is the hypotenuse (and $a,b$ are the legs), then we have that $12 c = ab$ (by equating the area of the triangle). $a+b+c$ and $c$ are both integers.
We also have that $(a+b)^2$ is a perfect square.
$$(a+b)^2 = a^2 + b^2 + 2ab = c^2 + 24c$$
By imagining a semicircle of diameter $c$, we can easily see that $c \ge 24$, and note that for any $c \ge 24$, a right triangle exists.
Now $c^2 + 24c$ has to a perfect square, and $c=25$ is the smallest that works.
The book had the answer as $24$ which is wrong.
