## Page 1 of 4

64 At Right Angles | Vol. 3, No. 1, March 2014 problem corner

Squares

between Squares

I. A ‘Trivial’ Observation

II. Applications

I. A ‘TRIVIAL’OBSERVATION

It is often the case — more often than one may

expect— that in the successful solution of a

problem, the seed of the solution is a very simple

observation; so simple that we call it ‘trivial’, and

the solution of any problem. We showcase such an

instance here, starting with the following:

Observation 1.

Between two consecutive integers there does not lie

another integer.

Stated this way the observation looks too trivial to

be of use to anyone. But even simple observations

have consequences! And these may well be of a

less trivial nature than the original one. They in

turn may give rise to other consequences, and so

on, and these may progressively get less trivial.

and far from trivial!

For example, here is a consequence of the

Observation 1, which we get by squaring the

quantities:

Observation 2.

Let be an integer. Between the perfect squares

and there does not lie another perfect

square.

From this we deduce the following, which looks

more impressive than Observation 1:

Observation 3.

Let and be integers such that

. Suppose that is a perfect

square. Then either or .

II. APPLICATIONS

We make use of observations 2 and 3 made above

in solving two problems posed in the ‘Adventures’

column of the November 2013 issue of AtRiA.

Find all integers such that is a perfect

square.

Solution Suppose that . Then

, so lies

between the two consecutive squares and

and thus cannot be a square. So there can

be no solution with .

Next suppose that ; then , or

.

1

I. A ‘TRIVIAL’OBSERVATION

It is often the case — more often than one may

expect— that in the successful solution of a

problem, the seed of the solution is a very simple

observation; so simple that we call it ‘trivial’, and

the solution of any problem. We showcase such an

instance here, starting with the following:

Observation 1.

Between two consecutive integers there does not lie

another integer.

Stated this way the observation looks too trivial to

be of use to anyone. But even simple observations

have consequences! And these may well be of a

less trivial nature than the original one. They in

turn may give rise to other consequences, and so

on, and these may progressively get less trivial.

and far from trivial!

For example, here is a consequence of the

Observation 1, which we get by squaring the

quantities:

Observation 2.

Let be an integer. Between the perfect squares

and there does not lie another perfect

square.

From this we deduce the following, which looks

more impressive than Observation 1:

Observation 3.

Let and be integers such that

. Suppose that is a perfect

square. Then either or .

II. APPLICATIONS

We make use of observations 2 and 3 made above

in solving two problems posed in the ‘Adventures’

column of the November 2013 issue of AtRiA.

Find all integers such that is a perfect

square.

Solution Suppose that . Then

, so lies

between the two consecutive squares and

and thus cannot be a square. So there can

be no solution with .

Next suppose that ; then , or

.

1 ## Page 2 of 4

Vol. 3, No. 1, March 2014 | At Right Angles 65

Hence nnnn� + 2nnnn + nn nn nnnn� + nnnn + nn nn nnnn�. So the only

way for nnnn� + nnnn + nn to be a square is to let

nnnn + nn nn nn, i.e., nnnn nn nnnn.

Since nnnn nn nn also yields a square value, the

possible nnnn-values are nnnn, nn. Both these nnnn-values

yield nnnn� + nnnn + nn nn nn�.

Problem 4, ‘Adventures’, November 2013 may

be solved in the same way.

Find all integers nnnn such that nnnn� + nnnn� + nnnn� + nnnn + nn is

a perfect square.

This is a substantially more complicated problem

of the same genre, but it too can be solved the

same way. The �irst challenge is to box the given

quantity between two perfect squares. To do this

we shall use the ‘completing the square’ reasoning

yet again.

Recall that to add a term to (xxxx� + aaaaxxxxaa so as to get a

square, we halve the coef�icient of xxxx and use that

to create the term: xxxx� + aaaaxxxx xx �

� aaaaaa� nn nnnnnn �

� aaaaa�

.

The same logic applied to nnnn� + nnnn� + nnnn� + nnnn + nn

suggests that the expression to be studied has to

be nnnn� + �

� nnnnn

. For, when we expand this

expression, we get both the nnnn� and nnnn� terms with

the correct coef�icients:

�nnnn� +

nnnn

2�

nn nnnn� + nnnn� +

nnnn�

4 .

Note that we have not got the right coef�icient for

nnnn� (we have got �

� instead of nn), and we haven't

got the other two terms at all.

The fractions ( �

� and �

� ) make things a bit

awkward, so why don't we multiply everything by

4? That way, squares remain squares (for, if XXXX is a

square, then so is 4XXXX; this is crucial), and at the

same time we get rid of the fractions. So we ask:

For which integers nnnn is the quantity

AAAA AAAAAAAA� + nnnn� + nnnn� + nnnn n nnn a square?

So let us now box AAAA between two consecutive

squares. Since 2 × nnnn� + �

� nnnnn nn nnnnn� + nnnn, the

candidate squares are AAAA AAAAAAAA� + nnnnn

� and

CCCC CCCCCCCC� + nnnn n nnn

. We now have:

2nnnn� + nnnnn

� nn 4nnnn� + 4nnnn� + nnnn�,

∴ AAAA nn AAAA nn AAnnnn� + 4nnnn + 4.

We shall show that this quantity is always positive.

There are various ways of seeing this. The

simplest is to compute the discriminant of the

quadratic expression AAnnnn� + 4nnnn + 4 using the

well-known “bbbb� nn 4aaaaaaaa” formula. We get:

4� nn 4 × AA × 4 nn nn6 nn 48 which is negative. As the

discriminant is negative, the expression

AAnnnn� + 4nnnn + 4 never changes sign. And since it is

positive for nnnn nn nn, it is positive for all nnnn.

(Another way to see that AAAA nn AAAA AA nn is to write the

expression for AAAA nn AAAA as:

AAAA nn AAAA nn AAnnnn� + 4nnnn + 4

nn nnnn� + nnnn� + nnnn� + 4nnnn nnn

nn nnnn� + nnnn� + (nnnn + 2aa�.

Please complete this line of reasoning on your

own.)

We see that AAAA AA AAAA, strictly. Now let us look at

CCCC nn nnnn:

2nnnn� + nnnn n nnn

� nn 4nnnn� + 4nnnn� + 5nnnn� + 2nnnn + nn,

∴ CCCC nn nnnn nn nnnn� nn 2nnnn nn AAA

Conveniently for us, the form nnnn� nn 2nnnn nn AA

factorizes:

nnnn�nn2nnnnnnAA nn (nnnnnnAAAAAnnnn+nnaaaaaaaannAAAA nn (nnnnnnAAAAAnnnn+nnaaa

The quadratic expression (xxxx nn nnnnnxxxx + nnnn gives rise

to the following graph:

Hence CCCC nn nnnn is positive if nnnn nn nnnn or if nnnn AA AA.

2

We see that the expression (n -3) (n +1) has the

following sign profile: ## Page 3 of 4

66 At Right Angles | Vol. 3, No. 1, March 2014

This has important consequences:

Proposition 1. If either nnnn nn nnnn or nnnn nn nn then

(��� � �)

� � � (�� � �� � �� � � � �) � (��� � � � �)

� .

So if the integer nnnn is less than nnnn or greater than nn,

the quantity 4 �nnnn� + nnnn� + nnnn� + nnnn n nnn lies strictly

between two consecutive squares and is thus not a

perfect square. Therefore the quantity

nnnn� + nnnn� + nnnn� + nnnn + nn is not a square if nnnn nn nnnn or

nnnn nn nn.

Proposition 2. If nnnn ≤ nnnn ≤ nn then

(��� � �)

� � � (�� � �� � �� � � � �) � (��� � � � �)

� .

Equality holds in the inequality on the right side

precisely when nnnn nn nnnnnnnn nnnn. Therefore, the quantity

4 �nnnn� + nnnn� + nnnn� + nnnn n nnn is a perfect square

precisely when nnnn nn nnnnnnnn nnnn.

We have fully solved the problem! The answer is:

The quantity nnnn� + nnnn� + nnnn� + nnnn + nn is a perfect

square precisely when nnnn nn nnnnnnnn nnnn.

All this from the ‘trivial’ observation that between

two consecutive integers there does not lie

another integer! We have come a long way indeed.

We invite you to apply these ideas to the following

problem:

Find all integers nnnn such that

nnnn� + 2nnnn� + nnnnnn� + 4nnnn + nn is a perfect square.

Remark 1. Here is a problem which closely

resembles the one solved in the preceding pages:

Find all integers nnnn such that nnnn� + nnnn� + nnnn + nn is a

perfect square.

But the resemblance is deceptive. Though the

problem concerns a polynomial of lower-degree,

it is signi�icantly more dif�icult than the one we

solved. Perhaps the main reason for this is that its

degree is odd (nn rather than 4), so there is no easy

way of boxing it between two squares. We shall

not attempt to solve the problem here.

Remark 2. We mention here a second

consequence of Observation 1 (but in another

form); it is based on Problem 11121 from the

well-known journal American Mathematical

Monthly. For any positive integer nnnn, we consider

the set SSSS� of integers that lie strictly between nnnn�

and (nnnn + nn)�. For example, SSSS� = nn2nn nnnn,

SSSS� = nnnnnn 6nn 7nn 8nn and SSSS� = nnnn0nn nnnnnn nn2nn nnnnnn nn4nn nnnnnn.

You can check that SSSS� has 2nnnn elements. We now

�an we �ind two distinct numbers in SSSS� whose

product is a perfect square?

The answer is clearly ‘No’ for nnnn = nn and nnnn = 2

for every nnnn. The proof is elegant and instructive.

Proof We adopt the well-known approach of

‘proof by contradiction’. Suppose that for some

nnnn nn nn there exist integers aaaann aaaa nn nnnn�, aaaa nn aaaa, such

that aaaaaaaa is a square. Now any positive integer can

be written as the product of a perfect square and a

square-free integer (i.e., an integer not divisible

by any perfect square larger than nn), simply by

factoring out the largest possible perfect square

from it. (For example, 20 = 2� × nn; nnnn0 = nn� × 6.)

Accordingly, we write aaaa = aaaaaaaa� and aaaa a bbbbbbbb�

where aaaa and bbbb are square-free. Since aaaaaaaa is a

perfect square, so is aaaaaaaa; but this implies that aaaa

and bbbb are the same number! (Do you see why? It

is because each prime factor in aaaa and bbbb occurs

just once in each number.)

So we have aaaa = aaaaaaaa� and aaaa a aaaabbbb� for some

positive integers aaaa, aaaa, bbbb with aaaa aaaaaa. It must be

that aaaa aaaa, else we get the chain

nnnn� nn nnnn� nn nnnn� nn (nnnn + nn)�, which is clearly absurd.

Now from the relation nnnn� nn nnnnnnnn� nn nnnnnnnn� nn (nnnn + nn)�

we get, by taking square roots:

nnnn

√aaaa

nn nnnn nn nnnn nn

nnnn + nn

√aaaa .

The difference between the numbers at the ends

of this chain of inequalities is

nnnn + nn

√aaaa nn nnnn

√aaaa = nn

√aaaa

nn nnnn

i.e., the difference is strictly less than nn. On the

other hand, bbbb bbbbbbbbbb: the difference between the

middle numbers is not less than nn. These relations

contradict each other! Hence such a situation

cannot happen. That is, integers aaaa, aaaa with the

stated property do not exist.

3 