# Given a right triangle whose side lengths are all integer multiples of 8, how many units are in the smallest possible perimeter of such a triangle?

Given a right triangle whose side lengths are all integer multiples of 8, how many units are in the smallest possible perimeter of such a triangle?

Does this mean the side lengths and not the hypotenuse or does it mean all three sides?

• All three sides, dear… – Parcly Taxel Feb 24 '18 at 3:36
• So if they're all integer multiples of 8, then the sides would have to be 24, 32, and 40. So P=96. – ddswsd Feb 24 '18 at 3:38

We have $$(8a)^2+(8b)^2=(8c)^2$$ Dividing by 64: $$a^2+b^2=c^2$$ As is well-known, the smallest possible integer values for these variables are $a=3,b=4,c=5$, so the smallest possible perimeter is $$8(a+b+c)=8\cdot12=96$$