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Choose $a_1,a_2,\dots,a_{50}$ from the set $\{1,2,3,\dots,98\}$ at random. Denote by $A$ the set that contains the chosen numbers. State, with arguments, if the followings assertion are true or false.

a) There exists two numbers in $A$ such that their sum is a cube, that is there exists $x$ and $y$ in $A$ such that $a+b=x^3$.

b) There exists at least one prime number in $A$.

c) The difference of any two numbers of $A$ is a perfect square.

d) The sum of any two numbers of $A$ is not a perfect square.

My attempt was to remove the items that it looks wrong. For instance, I think that there exists a list of $50$ numbers from $\{1,2,...,98\}$ such that c, d, b does not hold. For instance:

  • Choose $68, 43, 93, 14, 10, 19, 50, 13, 77, 66, 5, 31, 34, 8, 39, 32, 75, 36, 22, 70, 17, 85, 46, 61, 87, 58, 53, 4, 26, 33, 79, 23, 62, 90, 65, 69, 81, 49, 78, 54, 20, 27, 64, 41, 12, 21, 82, 1, 15, 72$
  • Then $43$ is prime, so b fails to be true.
  • Then $1+15=16$ is a perfect square, so d fails to be true.
  • Then $14-10=4$ is perfect square, so c fails to be true. What about a? I don't see how to check it? I am missing something? Thank you in advance.
roydiptajit
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stefano
  • 625
  • @Babado Are you sure your edit phrased (c), (d) as intended? The original is ambiguous whether it is a "for all" or "there exists" but given the title "pigeohole" it is probably "for all $A$ there exists two members of $A$ such that ...". – user10354138 Feb 27 '21 at 12:46
  • I thought it was meant that, but please @stefano correct me if that is not the correct statement. – Babado Feb 27 '21 at 12:50
  • @Babado Then let's write "there exist"? Or to be less formal, it should at least be "of some two numbers", not "of any two numbers". – aschepler Feb 27 '21 at 12:54
  • I think this is four different questions to be answered independently, not asking if a single set satisfies all four conditions or none of the four conditions. In your example, b, c, and d are all true, not false - but that doesn't answer whether any of the conditions is true for every possible choice of the set. – aschepler Feb 27 '21 at 12:59
  • English is not my native language, but I try to type a better formulation for c. The original question asks to state that is true or false the following assertion: if we choose arbitrarily 2 of 50 numbers, then their sum is a perfect square. Similar for c. I am not sure that it can be used the piegonhole principle, it is just a guessing since I've studied about this principle. – stefano Feb 27 '21 at 13:02
  • for c), by the pigeonhole, there are at least two numbers which difference is $1$, that is, there is at least a pair of numbers which difference is a perfect square. – Babado Feb 27 '21 at 13:03
  • for b), between $1$ and $98$, there are $25$ primes, so you can just chose $50$ composite numbers for the set $A$. – Babado Feb 27 '21 at 13:05
  • @aschepler So, b is true? Why? – stefano Feb 27 '21 at 13:44
  • No, b is false. You can pick $50$ numbers between $1$ and $98$ such that none of them is prime. Therefore picking any number from that set, you will never get a prime. (I think this is what you meant). – Babado Feb 27 '21 at 14:03

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