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This work is inspired in part by a funny math video on Youtube in which Dany, a young student insisted that \(2 + 2 =22\), despite objections from his teacher. The other part of inspiration came from a comment of the Indian Prime Minister Narendra Modi, who declared in an election speech that 'will make \(1+1=11\)'. In this talk, we will ask a more general question: when does the group law '\(+\)' satisfy both \(1+1=u\) and \(2+2=v\)? Surprisingly, this innocuous and perhaps strange looking question has connections with some interesting results in number theory. We will talk about such connections and in particular refer to works of ancient Indian mathematician Brahmagupta, who in early 6th century developed a method to solve Pell's equation (i.e., Diophantine equation of the form \(x^2 - ny^2 =1\), where \(n\) is a positive square-free integer). We will show that if \(|u-v+1|= 1\), then \(1 +1 = u\) and \(2 + 2 = v\) (with \(+\) defined by a polynomial function) is always possible in the field of rational numbers \(\mathbb{Q}\). On the other hand, given integers \(u\), \(v\), and \(t\) such that \((u-1)(v-2) = 2 t^2\), then \(1 + 1 = u\) and \(2 +2 = v\) (with \(+\) defined by a polynomial function) is possible in the field of rational numbers \(\mathbb{Q}\) , if and only if \(t = [(3 +2 \sqrt(2))^m - (3 -2\sqrt(2))^m ] / (4 \sqrt(2)\) for some non-negative integer \(m\).

References:

1. R. Padmanabhan, Group laws satisfying \( 1 + 1 =11\) and \(2 + 2 =22\) , this Seminar, January 29, 2019, abstract