Let AB be a leg of the right triangle of least perimeter whose sides have integral lengths, whose hypotenuse is one unit longer than AB, and in which AB > 100. Find AB.

bigmac Dec 8, 2019

#1**+1 **

*Let AB be a leg of the right triangle of least perimeter whose sides have integral lengths, whose hypotenuse is one unit longer than AB, and in which AB > 100. Find AB.*

Let's call the sides a, b, and c, for my convenience. Let AB be a, the other leg b, and the hypotenuse c.

By Pythagoras' Theorem a^{2} + b^{2} = c^{2}

a must be larger than 100

c must be one unit larger than a

all values must be integers

I don't have a general proof for this;

I had to do it by trial and error. AB (that is, a) equals **112**

The three sides of the triangle are 112, 15, and 113

112^{2} + 15^{2} = 113^{2}

Since AB had to be larger than 100, I started with 101, squared it, subtracted that from 102 squared, and checked if the remainder were a perfect square. It wasn't. I moved on to 103^{2} minus 102^{2} and the remainder was not a perfect square. When I tried 104^{2} minus 103^{2} I noticed a pattern to the remainders so I skipped ahead to 111^{2} minus 110^{2} and that wasn't it either but I was approaching 225. I finally found it at 113^{2} minus 112^{2} .... whew.

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Guest Dec 8, 2019