A collection of interesting and challenging math problems. Click “Show Solution” to reveal the answer.
System of Equations (Attributed to Srinivasa Ramanujan)
\(x,y\in\mathbb{Z}\). Solve for \(x\) and \(y\):
$$\sqrt{x}+y=7$$
$$x+\sqrt{y}=11$$
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Let \(p=\sqrt{x}\,\) and \(q=\sqrt{y}\)
$$p+q^2=7$$
$$p^2+q=11$$
$$11=7+4=p+q^2+4$$
so
$$p^2+q=p+q^2+4$$
Rearrange
$$p^2-q^2+q-p=4$$
Since
$$p^2-q^2=(p-q)(p+q)$$
Substitute
$$(p-q)(p+q)+q-p=4$$
$$(p-q)(p+q)-(p-q)=4$$
$$(p-q)(p+q-1)=4$$
There are three ways to factor \(4\): \(2\times 2, 4\times 1, and 1\times 4\).
Assume \(p-q=2\) and \(p+q-1=2\).
$$p-q=2$$
$$p+q-1=2$$
Add the two equations.
$$2p-1=4$$
$$2p=5$$
$$p=5/2$$
This is not an integer, so not a valid solution.
Assume \(p-q=4\) and \(p+q-1=1\).
$$p-q=4$$
$$p+q-1=1$$
Add the two equations.
$$2p-1=5$$
$$2p=6$$
$$p=3$$
$$3-q=4$$
$$-q=1$$
$$q=-1$$
\(\sqrt{y}\ne -1\), so this is not valid.
Assume \(p-q=1\) and \(p+q-1=4\).
$$p-q=1$$
$$p+q-1=4$$
Add the two equations.
$$2p-1=5$$
$$2p=6$$
$$p=3$$
$$3-q=1$$
$$-q=-2$$
$$q=2$$
Thus
\(x=p^2=9\) and \(y=q^2=4\)