Almost a Pythagorean Triple

Number Theory Level 4

You're given a right triangle as shown above. If a a is a positive integer such that it satisfies the constraints a ∣ c a|c and b 2 = a ! + ( a − 1 ) ! + ( a − 2 ) ! b^{2} = a! + (a-1)! + (a-2)! , what is the largest known value of a a ?

Hint : The related problem remains one of the world's unsolved math mysteries.


The answer is 9.

Worranat Pakornrat by Worranat Pakornrat , 36, Thailand

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1 solution

From the Pythagorean theorem, a 2 a^{2} + b 2 b^{2} = c 2 c^{2} .

Then b 2 = a ! + ( a − 1 ) ! + ( a − 2 ) ! = [ ( a − 2 ) ! ] ( a ( a − 1 ) + ( a − 1 ) + 1 ) = [ ( a − 2 ) ! ] ( a 2 b^{2} = a! + (a-1)! + (a-2)! = [(a-2)!](a(a-1) + (a-1) +1) = [(a-2)!](a^{2} ).

Thus, a 2 a^{2} + b 2 b^{2} = a 2 a^{2} + [ a − 2 ) ! ] ( a 2 a-2)!](a^{2} ) = c 2 c^{2}

Since a ∣ c a|c , c 2 c^{2} = a 2 a^{2} d 2 d^{2} for some integer d.

Dividing a 2 a^{2} both sides, we will get: 1 + ( a − 2 ) ! (a-2)! = d 2 d^{2} .

This equation leads to the famous Brocard's problem: find the natural number n such that n!+1 is a perfect square.

Until now, there are only 3 known values for n: 4, 5, and 7.

4! + 1 = 25 = 5^2.

5! + 1 = 121 = 11^2.

7! + 1 = 5041 = 71^2.

By checking up to a billion, no other larger integers satisfy the condition, and many believe 7 is the highest number despite having no standard proofs yet.

As a result, since a − 2 a-2 = 7, the maximum known value of a a is 9.

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