# True and False with A Difference of Two Squares Dilemma

Question in Image Below

Solution

22) Fact

Perfect Squares can only have last digits 0, 1, 4, 5, 6, 9

Perfect Square CANNOT have these last digits: 2, 3, 7, 8

If (i) and (ii) were true, a number with last digit 2 is a perfect square, impossible!

If (ii) and (iii) are true, a number with last digit 1 – 48 has last digit 3 is a perfect square, contradiction again!

The only possibility is that statement (i) and (iii) must both be true

Statement (ii) is false

A + 41 = x^2   and A – 48 = y^2

Difference

x^2 – y^2 = 89 , x > y

(x – y)(x + y) = 89 and 89 is a prime number

(x – y) = smaller factor = 1

and

(x + y) = 89

(x – y) + (x + y) = 1 + 89

2x = 90

x = 45

Put back in (x – y) = 1

45 – y = 1

y = 44

Finally

A + 41 = x^2

Becomes

A + 41 = 45^2

A = 1984

Check

(i) A + 41 is a Perfect Square

1984 + 41 = 2025 = 45^2 (True)

(ii) Last digit of A = 1984 is 1 (False)

(iii) A – 48 is a Perfect Square

1984 – 48 = 1936 = 44^2 (True)