## 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

outtimedAugust 27, 2008 at 12:45 pm

very nice concept and lucid too..

ShivaniAugust 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

jd2718August 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? « JD2718August 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

CluelessSeptember 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?

PradeeSeptember 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 DwivediMarch 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.

IndraApril 24, 2009 at 7:07 am

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..

Thanks

HS

hsJuly 19, 2009 at 7:07 pm

k^2=n(n+16)

now n is positive so n(n+16) is positive hence k is positive..

Regards

Rahul

RahulJuly 23, 2009 at 8:12 am

rahul are u fool?

k = + or –

saurabhNovember 26, 2009 at 11:58 am

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..

Thanks

HS

hsJuly 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

RahulJuly 24, 2009 at 6:57 am

Excellent work!

:applause: 🙂

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