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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.

 Dec 8, 2019
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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                                 a2 + b2 = c2               

                                                                        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     

                                                                          1122 + 152 = 1132      

 

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 1032 minus 1022 and the remainder was not a perfect square.  When I tried 1042 minus 1032 I noticed a pattern to the remainders so I skipped ahead to 1112 minus 1102 and that wasn't it either but I was approaching 225.  I finally found it at 1132 minus 1122 .... whew.

.

 Dec 8, 2019

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