Concept 1 Perfect Squares
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
(x^3+y)(x^3-y)=127
so clearly 2x^3=128 x=4 and y=63
so one pair (4,63)
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
4m^2+100m+25^2-4k^2=25^2-76
(2m+25)^2-(2k)^2=549
(2m+25+2k)(2m+25-2k)=549.1, 183.3,61.9
4m+50=550,186,70
m=125,34,5
so sum is 125+34+5=164
outtimed
August 27, 2008 at 12:45 pm
very nice concept and lucid too..
Shivani
August 28, 2008 at 12:36 pm
Very nice. And not such a hard reach for students who have some fluency with factoring. Thank you.
Jonathan
jd2718
August 31, 2008 at 3:00 pm
[...] Unrelated, interesting note: Nice factoring techniques for solving problems such as Find n such that n(n+16) is a perfect square [...]
Number Puzzle: Who could I be? « JD2718
August 31, 2008 at 3:48 pm
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
Clueless
Clueless
September 2, 2008 at 1:56 pm
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
[(m+1)(m+24)]-5=k^2
[(m+1)(m+24)]-k^2=5
(m+1-k)(m+24+k)=5
The sum of these two equations will be 2m+25. So it will be an odd number. How to proceed from here?
Pradee
September 29, 2008 at 7:27 am
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.
Thanks!!!
Rahul.
Rahul Dwivedi
March 26, 2009 at 10:41 am
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.
Indra
April 24, 2009 at 7:07 am