Question

How many Pythagorean triangles are there with integer sides and one leg in Ramanujan number 1729?
A. 10
B. 11
C. 12
D. 13
E. None.

Answer 1

Hint: For solving this question you should know about the Pythagorean triple which consists of three positive integers a, b and c in such a way that it will be $a^2+b^2=c^2$. The Ramanujan number 1729 is the sum of cubes of sides of a Pythagorean triple or triangle.

Complete step-by-step solution:
According to the question we have to find the number of Pythagorean triangles which are there with integer sides and whose one leg is the Ramanujan number. According to Ramanujan theory, it is clear that 1729 is the sum of two different cubes, so if we take a cube of 10, then it is 1000 and if we take a cube of 9, then it is 729. So, the total of both are 1729 and that is Ramanujan number. So, according to our question it is clear that the triple’s Pythagoras equation is,
${1729}^2+b^2=c^2$
Or, we can say,
$\begin{array}{rl}& c^2-b^2={\left(1729\right)}^2\\ & \Rightarrow c^2-b^2=2989441\end{array}$
And 2989441 has 13 pairs of factors which are all different and they finally make a Pythagorean triple which has a Ramanujan number leg. Some of them are,
First triple: ${\left(672\right)}^2-{\left(1855\right)}^2={\left(1729\right)}^2$
Second triple: ${\left(1140\right)}^2-{\left(2071\right)}^2={\left(1729\right)}^2$
Third triple: ${\left(2665\right)}^2-{\left(2028\right)}^2={\left(1729\right)}^2$
Fourth triple: ${\left(4321\right)}^2-{\left(3960\right)}^2={\left(1729\right)}^2$
Fifth triple: ${\left(6175\right)}^2-{\left(5928\right)}^2={\left(1729\right)}^2$
So, 13 triples satisfy this and hence the correct answer is option (D).

Note: We should calculate the Ramanujan number leg in the triple’s Pythagoras theorem or equation. We have to calculate this on the big digits or greater digits because every number has a lot of difference and it can be done by mathematical calculations only.