Let A ⊆ {1, 2, . . . , 12} with |A| = 5. Prove that there are (a, b),(c, d) ∈ A × A with (a, b) /= (c, d) and a + b = c + d. I know that this requires the use of the pigeon rule but I am a bit confused about how that works or how to apply it here.
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Maybe consider the differences between the elements of $A$? What can you say about the differences between $a, b, c, d$ if $a+b = c+d$? – Brian Tung Apr 23 '20 at 23:04
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Ask yourself the following questions. How many pairs do you have in $A\times A$? How many distinct possibilities do you have for the sum of coordinates of pairs in $A\times A$? Is it possible for $(a, b)$ and $(b, a)$ to have distinct sums? These should lead you to the answer. – Besfort Apr 23 '20 at 23:05
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Clearly the pairs (a,b) (b,a) work. Are you perhaps asking for when $a\neq b\neq c \neq d$? – Alex R. Apr 23 '20 at 23:11
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Here's a MathJax tutorial :) – Shaun Apr 24 '20 at 00:41