# Form a Pythagorean triple in which the first integer is 11. Please show your workings clearly.Thanks

### 1 Answer | Add Yours

Pythagorean triples can be formed using two integers m,n m>n using Euclid's formula: If `a^2+b^2=c^2` then `a=m^2-n^2,b=2mn,c=m^2+n^2`

If 11 is to be the first number, then `m^2-n^2=11`

so we need n such that `sqrt(11+n^2)` is an integer. Trying a few values we find n=5 works: `sqrt(11+5^2)=sqrt(36)=6`

So m=6 and n=5 generating the triple:

`6^2-5^2=11,2(6)(5)=60,6^2+5^2=61`

**One triple is 11,60,61**