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Concept 1 Perfect Squares

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Concept I Perfect Squares

There has been a huge surge in the number of questions about perfect squares, in almost all mocks. The basic trick to any such question is assuming the number as a perfect square of an integer k and then using techniques of completion of square and then the formula of (a^2-b^2) and solving using divisibility theory

Example I Find all natural n such that n(n+16) is a perfect square

step 1 n(n+16)=k^2

step 2 (n^2+2.8.n+8^2)-8^2=k^2

step3 (n+8+k)(n+8-k)=64

see now lhs and rhs both are integers then both of (n+8-k) and (n+8+k) are divisors of 64. But note that we add the two equations we will get 2n+16, so teh sum of two divisors should be even hence both divisors even or both odd

so n+8+k=32,16,8,4,2 and n+8-k=2,4,8,16,32

but see this n is positive hence k is positive, thus n+8+k>n+8-k
so only two options

and solving we get 2n+16=34,20
so n=9,2

Note : The source of this problem is Pomona Wisconsin mathematics talent search exam!

Practice problem!!
Find the sum of all such positive integers m’s such that m^2+25m+19 is a perfect square

Now we will extend the method to other kinds of problems
Basically what we used in the above problem is difference of square method

lets take an example
x^6=y^2+127, find the no of pairs of postive integers (x,y)

first step in this problem is recognising that 127 is a prime
then we move to
so clearly 2x^3=128 x=4 and y=63

so one pair (4,63)

Written by Implex

August 25, 2008 at 5:18 pm

Posted in concepts, Number Thoery

Tagged with

23 Responses

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  1. Solution to Practice problem!!
    Find the sum of all such positive integers m’s such that m^2+25m+19 is a perfect square

    m^2+25m+19 =k^2
    m^2+2.(25/2)m+ (25/2)^2+19-(25/2)^2=k^2
    (2m+25+2k)(2m+25-2k)=549.1, 183.3,61.9

    so sum is 125+34+5=164


    August 27, 2008 at 12:45 pm

  2. very nice concept and lucid too..


    August 28, 2008 at 12:36 pm

  3. Very nice. And not such a hard reach for students who have some fluency with factoring. Thank you.



    August 31, 2008 at 3:00 pm

  4. […] Unrelated, interesting note: Nice factoring techniques for solving problems such as Find n such that n(n+16) is a perfect square […]

  5. Find all n such that n(n+16) is a perfect square:

    Unless the notation means that ‘n’ is a positive integer, there are other solutions: n=0, n=-18, n=-27



    September 2, 2008 at 1:56 pm

  6. Hi,

    For the practise problem I did like this. Please tell whether I am right and how to proceed

    (m^2+25m+(19+5))-5 = k^2
    The sum of these two equations will be 2m+25. So it will be an odd number. How to proceed from here?


    September 29, 2008 at 7:27 am

  7. hi pradeep,
    That was a nice innovative way of proceeding for the solution but I think u have made a small mistake there.
    The derivation of the step from [(m+1)(m+24)]-k^2=5 to (m+1-k)(m+24+k)=5 is incorrect because A^2 – B^2 = (A+B)(A-B).
    B^2 is k^2 whats A^2 here?

    So u need to convert LHS Into the form A^2-B^2. See the solution explained at the top of all comments.

    In case u have got the solution proceeding in your way. Please update the same to me.


    Rahul Dwivedi

    March 26, 2009 at 10:41 am

  8. Hi Rahul,
    In outtimed’s solution he/she has considered 1 as odd.Is that true?Isn’t 1 unique.If that is the case then the sum will be less by 125 i think.


    April 24, 2009 at 7:07 am

  9. Hi,
    I have a little problem in this step..dint understand it fully..
    “but see this n is positive hence k is positive, thus n+8+k>n+8-k”
    my problem is if n is positive how come it can be inferred that k would also be positive..
    as u have written in previous steps..
    n+8+k= 32,16,8,4,2
    taking, n+8+k =2 , implies n+k = -6
    since n is a natural number, it only means k should be a negative number..
    I then thought for a while and this thing came in my mind:-
    sum of the two factors is 2n+16..
    and since n >= 1
    so 2n+16>= 18
    which gives us only two only values of 2n+16: 34, 20..

    i don’t know how correct I am…
    please correct me..



    July 19, 2009 at 7:07 pm

  10. k^2=n(n+16)
    now n is positive so n(n+16) is positive hence k is positive..



    July 23, 2009 at 8:12 am

    • rahul are u fool?
      k = + or –


      November 26, 2009 at 11:58 am

  11. Hi Rahul,
    I am so sorry to cause this annoyance.
    Still, I feel like I am missing something..
    you said “k^2 = n(n+16)..
    now n is positive so n(n+16) is positive hence k is positive..”

    lets say, n=2 (taking the value of n from solution)..
    which gives
    k^2 = 2*18 = 36
    or k = 6 and -6

    Please correct me..



    July 23, 2009 at 7:55 pm

    • we are taking perfect squares, which means squares of positive integers.

      there is no use taking into account negative intgeres, they will give unwanted roots


      July 24, 2009 at 6:57 am

  12. Excellent work!


    August 17, 2009 at 3:23 pm

  13. Really a needful blog. Excellent way to follow these kind of problems. I use to take these as time consuming sums and skip it. I wont skip from now. Thanks


    August 20, 2009 at 11:48 am

  14. Hey grttt work thnks a lott


    September 20, 2009 at 4:56 am

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    September 20, 2009 at 4:57 am

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