The sides of a right angled triangle are all integers.Two sides are primes that differ by 50.The smallest possible value of the third side is ............

• 60
• 57
• 53
• 49

Let the two sides be a and b ,where a is prime,
Since,the third side differ by a by 50
so the third side = 50 + a.

Now, by pythagoras theorem,
  $$\begin{gathered} a^2 + b^{2 } = (a+50{)}^2 \\ \Rightarrow a^2 + b^{2 } = a^2 + 100a +2500 \\ \Rightarrow b^{2 }= 100 a + 2500 \end{gathered}$$
  Now,the minimum value of a to satisfy the equation and b to be an integer,is 11
 
i,e,  $$\begin{gathered} b^2=100\times 11 + 2500 \\ \Rightarrow b = \sqrt{3600} = 60 \end{gathered}$$
Thus,the sides of the triangle are 11,60,61

Thus,the smallest possible value of third side is 60.