## Le Monde puzzle [#839]

**A** number theory Le Monde mathematical puzzle whose R coding is not really worth it *(and which rings a bell of a similar puzzle in the past, puzzle I cannot trace…)*:

The set Ξ is made of pairs of integers (x,y) such that (i) both x and y are written as a sum of two squared integers (i.e., are bisquare numbers) and (ii) both xy and (x+y) are bisquare numbers. Why is the product condition superfluous? For which values of (a,b) is the pair (13^{a},13^{b}) inΞ?

**I**n the first question, the property follows from the fact that the product of two bisquare numbers is again a bisquare number, thank to the remarkable identity

(a²+b²)(c²+d²) = (ac+bd)²+(ad-bc)²

(since the double products cancel). For the second question, once I realised that

13=2²+3²

it followed that any number 13^{a} was the sum of two squares, hence a bisquare number, and thus that the only remaining constraint was that (b≥a)

13^{a+}13^{b}=13^{a}(1+13^{b-a})

is also bisquare. If b-a is *even*, this sum is then the product of two bisquare numbers and hence a bisquare number. If b-a is *odd*, I do not have a general argument to bar the case (it certainly does not work for 13+13² and the four next ones).

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