Consider the right angled triangle with sides $a=7$, $b=24$ and $c=25$.
The area of this triangle is $84$, which is divisible by the perfect numbers $6$ and $28$.
Moreover it is a primitive right angled triangle as $\gcd(a,b)=1$ and $\gcd(b,c)=1$.
Also $c$ is a perfect square.
We will call a right angled triangle perfect if
-it is a primitive right angled triangle
-its hypotenuse is a perfect square.
We will call a right angled triangle super-perfect if
-it is a perfect right angled triangle and
-its area is a multiple of the perfect numbers $6$ and $28$.
How many perfect right-angled triangles with $c \le 10^{16}$ exist that are not super-perfect?
Firstly, let’s define the terms:
Primitive right angled triangle: A right-angled triangle is said to be primitive if the greatest common divisor (gcd) of its sides is 1. This means that the three sides do not share any common factor other than 1.
Perfect square: A perfect square is an integer that is the square of an integer. For example, the numbers 4, 9, 16, 25, etc. are perfect squares.
Perfect number: A perfect number is a positive integer that is equal to the sum of its positive divisors excluding the number itself. For instance, 6 and 28 are perfect numbers since 1+2+3 = 6 and 1+2+4+7+14 = 28.
In this problem, we are asked to find the “perfect” right-angled triangles whose hypotenuse ‘c’ is a perfect square (up to 10^16) and are also primitive, but not “super-perfect” (i.e., their area is not a multiple of the perfect numbers 6 and 28).
To begin with, the Pythagorean theorem states that a^2 + b^2 = c^2, where ‘c’ is the length of the hypotenuse (the side opposite right angle), and ‘a’ and ‘b’ are the lengths of the other two sides. We use this theorem for right-angled triangles, since the question pertains to such triangles.
Recall that the area of a right triangle is (1/2)*a*b, and we want it NOT to be divisible by 6 and 28. Since 6 and 28 have a gcd of 2, we essentially need the area NOT to be divisible by 2*3*2*7 = 84.
The characters ‘a’, ‘b’ and ‘c’ represent lengths of sides. Since ‘c’ is the hypotenuse and is a perfect square up to 10^16, ‘c’ is a square of an integer ‘m’, which can go up to sqrt(10^16) = 10^8.
Note that when ‘m’ is odd, ‘a’ and ‘b’ are also odd and so the area becomes divisible by 1*1 = 1 but not 84. Hence, for every odd ‘m’ in the interval [1, 10^8], we get an odd ‘c’, and thus a perfect right-angled triangle that is not super-perfect.
When ‘m’ is even, ‘a’ and ‘b’ are of opposite parity (one is odd, the other is even). Therefore, the area could potentially be divisible by 2 but not 84.
However, due to the uniqueness in the primitive Pythagorean triple ‘a’, ‘b’, ‘c’ (where ‘a’=m^2-n^2, ‘b’=2*m*n, ‘c’=m^2+n^2 for some integers ‘m’ and ‘n’ with ‘m’>n), it will turn out that the area is always divisible by 84 when ‘m’ is multiple of 3 or 7. Therefore, it is enough to count only those even ‘m’ that are not multiples of 3 or 7.
So, the answer to the problem would be:
Number of odd ‘m’ up to 10^8 + Number of even ‘m’ up to 10^8 that are not multiples of 3 or 7.
To get these numbers, we use the inclusion-exclusion principle:
Number of odd ‘m’ up to 10^8 = (10^8 + 1) / 2 = 50,000,000.
Number of even ‘m’ up to 10^8 that are not multiples of 3 or 7
= (10^8 / 2) – (10^8 / 6) – (10^8 / 14) + (10^8 / 42)
= 50,000,000 – 16,666,667 – 7,142,857 + 2,380,952
= 28,571,428.
Thus, adding these two results we get the total number of perfect right-angled triangles with c <= 10^16 that are not super-perfect = 50,000,000 + 28,571,428 = 78,571,428.
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