Sarvesh Iyer and his great mathematical exploits

Life is great when you have something to hang around with.And for me,that is what numbers are like.They're fascinating,tell more than they can,and also are also hiding around us when we are unaware of it.Fibonacci numbers are all around us.In flowers,in superstition,in common devices,and interwoven in the extremely efficient design of nature.

That series goes somewhat like this:0,1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597 and so on.

And there is also the constant that is an epitome of beauty:the golden ratio,phi,equal to 1.618.

The Fibonacci series and the golden ratio are actually related!

​​The relation is called Binet's formula​. ​​​

Anyway,now I'll come to my work.

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I'll first tell you about my work on squares having only three digits,or less, in them.

Three/Two Digitized Squares

I call these squares as Three digitized,or Two digitized squares.These are squares that contain only 3 digits in them.Take for example,169,the square of 13.It has only 3 digits,all of which are different,so it falls in this category.Bigger examples include 40401,the square of 201,and 9559559529,the square of 97773.There are also some two digitized squares,like 144,which is 12 square,or 69696,the square of 264,or 6661661161,the square of 81619.​

So this is what I did:I wrote a program,in which I found every single square below 10^13,which comprised of only 3 digits or less.There are around 850 such squares in total.Some very exciting patterns emerge through this,but one of the most exciting things is that there are exactly two combinations of 3 distinct digits for which,given any number n between 1 and 12,we can find a number that has n digits, and contains only these digits. These combinations are (1,2,5) and (4,2,5).Something tells me that one of these is bound to carry on for a long time,while the other one will terminate at around 17 or 18 digits,but I can't say as yet.

​My next piece of work came on reading a book on certain type of numbers,which were in a book written by Kaprekar.Unfortunately, I forgot the name of these numbers,but I call them unnamed for now.

Unnamed numbers​

I define these numbers as those whose square,when written in base 10,can be completely digitally broken up into other perfect squares.Some small examples are 13, as 13 square=169="16"+"9", and 37,since 37 square=1369="1"+"36"+"9".Slightly bigger examples are 475(225625="225"+"625") and 112(12544="1"+"25"+"4"+"4").The best(not biggest) I've found is a 6 digit number,387288, whose square is very interesting as it contains only 1,4 and 9 as digits,so each digit is a perfect square."0" for my own reasons,is not defined as a perfect square.Oh,387288 square is 149991994944.This number can also be broken up into squares in more than one way;the 49 part can be broken up as "4" and "9" or just simply "49".Similarly, in the case of 580079,it's square is 336491646241.Here,3364(58 square) and 6241(79 square) are anyway squares,so we leave them out,and focus on 9164.9164 is "9" + "16" + "4",as well as "9" + "1" + "64".So again we have two possibilities for the same number.Another interesting thing is a pattern which immediately leads me to a lot of these numbers.Starting from 660033, we increase the last two digits as a single number by 1,and the first two by 2, to get 680034,and continue till 980049.Astonishingly,one will wonder why all these numbers are "unnamed".The proof is simple. If you square this number,treating it as say (680000+34)^2 and expanding by the common algebraic expansion,the resulting 12 digit number will have it's first 4 digits a perfect square,that of 680000,then the last four a perfect square of 34,and the middle four a perfect square of 6800!The reason being that the middle term is the value of 2ab in the algebraic expansion,and so it is 2*680000*34,which is 6800*6800,because 34 is half of 68(which explains the choice of our pattern).

Next I go on to explain what are automorphic numbers,and how these can be found out.

Automorphic numbers​

​These numbers are very interesting  in that their square ends digitally in themselves.For example,25 square is 625 and 625 ends in 25.Of course,there are slightly bigger examples in 376(square is 141376) and 625 itself!(390625 is it's square).And there are huge examples,like 8212890625(whose square is 67451572418212890625).Now I cannot say exactly why,but I  thought that maybe automorphic numbers could be generated by prefixing either a string of digits or a single digit itself to a smaller automorphic number.And that is precisely why I managed to get out many more automorphic numbers,up till 10^31.It's quite easy to find it,if you know the smaller examples of 5 and 6.

Say you want to start from 5.So prefix a random digit to 5,and find out if it is automorphic.If it is,then that is great.Otherwise,pick the integer either one less or one more than the earlier digit you chose.Observe an arithmetic trend in the digit,which comes in the place value immediately after 5.In our case,there is no trend,that digit is constant at 2.So we say 25 square=625 is automorphic. So there.

​And if we want to continue from 25,we do the same thing.Try 325.It's square is 105625.So 325 is not automorphic.Try 425.It's square is 180625.Note that the digit immediately after 25 in the square,6,has not changed.So we try 625 square to get 390625,and we've done it again!

Now try the same with 6.Say we take 16.16 square is 256.So 16 is not automorphic.We take 26.26 square is 676.So we notice that the digit in tens place after the 6 has increased by 2.Since the digit cannot exceed 10 or go below 0,it has to wrap around.So it would go 5,7,9,1,3 and back.

So we map the prefix digit to the changing digit,and the mapping precisely is:1->5,2->7,3->9,4->1,5->3,6->5,7->7,8->9,and 9->1.So we see that the only matching pair is 7 and 7,so that 76 should be automorphic.Indeed,it is.

But there can be a loop hole.What if there is no match?It is possible,for if you try extending 625,we don't get any matching mappings.

In that case,we add a 0 as prefix and proceed as normal.

Try that for 625(First try it out without the prefix,to assure yourself that there are no matches).Make it 0625,and try adding 1 as a prefix.10625 square is 112890625,so it is not automorphic.Try 20625.It's square is  425390625.We notice that the number 9 is fixed in the place immediately after the 0625 sequence.So we try 90625,and see that it is indeed automorphic.

So discover some more yourselves!

My next topic is even more interesting.It talks about perfect squares which are in arithmetic progression.

Squares in arithmetic progression

​An arithmetic progression is a series of numbers,which are either increasing or decreasing  by a constant figure.The option of squares in arithmetic progression is interesting,because examples can be hard to find by hand.A little experimentation can help a lot,though.

And so I did experiment,and found an interesting recurrence relation.

If a^2,b^2 and c^2 are in arithmetic progression,then if we define d=b,f=c and

e=sqrt(2((a-b)^2)+c^2)-a-b, then even d^2,e^2 and f^2 are in arithmetic progression in that order.(sqrt means the square root of whatever is in the brackets)

So we find the smallest a,b,c and implement the formula,provided e is an integer.

Our first triple is 1,5,7.Try it out.

Try these in the formula,and we have another triple:17,13 and 7.And putting these back in,we have 17,25,31.

Now imagine we list these down as a table.  

a 1  7    17  31  49  71  97   

b 5  13  25  41  61  85  113

c 7  17  31  49  71  97  127

Notice the pattern.The c of a triple becomes a of the next triple.The difference between b and a in each triple is increasing by 2,starting with 4 for the first triplet.The difference between c and b is also increasing by 2,and starts with 2 from the first triplet.So now,there is no need to even look back to the formula.Just enjoy filling this table!

Infact, it is well known that no four numbers in arithmetic progression can all be squares. The proof of this is quite long to be contained within these margins, however.