Welcome to Atharv’s Maths Blog! Today I will talk about Pythagoras’ theorem. If you don’t know, Pythagoras of Samos (550-495 BC) was a Greek thinker and the founder of the Pythagorean group. He formulated a theorem stating that \(a^2+b^2=c^2\) for only specific values of \(a\), \(b\) and \(c\). This theorem has over 200 proofs, but what we are most interested in for this post is how to generate those values of \(a\), \(b\) and \(c\). Let us take two integers \(m\) and \(n\) where \(m \ge 0\), \(n \ge 0\) and \(m \ge n\). Now the values of \(a\), \(b\) and \(c\) are given below. $$a = m^2 – n^2$$ $$b = 2mn$$ $$c = m^2+n^2$$ Very interesting… Is that because \(b + c = (m+n)^2\)?
Yes! \(b+c\) works out to \(m^2+n^2+2mn\), which, because of the famous algebraic identity, is \((m+n)^2\). So let us look at some examples. So if \(m=9\) and \(n=5\). Then the values of \(a\), \(b\) and \(c\) will be defined as follows: $$a=9^2 – 5^2=81-25=56$$ $$b=2\times9\times5=18\times5=90$$ $$c=9^2+5^2=81+25=106$$ So, as we can see, the triplets (Pythagoras’ triples or triplets are the values of \(a\), \(b\) and \(c\)) are 56, 90 and 106. And, the square of the first number plus the square of the second number is equal to the square of the third number! $$56^2+90^2=3136+8100=11236=106^2$$ Another example! If \(m=6\) and \(n=3\), then $$a=6^2-3^2=36-9=27$$ $$b=2\times6\times3=12\times3=36$$ $$c=6^2+3^2=36+9=45$$ The triplets in this case are 27, 36 and 45. And, 27 squared + 36 squared is indeed 45 squared! $$27^2+36^2=729+1296=2025=45^2$$ So now we know how to generate Pythagoras’ triplets. I hope I was able to make it interesting.
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This is the end of my first mathematical blog post. Have a nice day! See you next time!