Question

the number of integers n<20 for which n²-3n+3 is a perfect square

Answer 1

The answer to your question can be done by putting all the values of n, But it will take us some time , So Here is a shortcut method to do it,

Let's Convert the given expression to something new, So that our work can be made easier,

Give Expression : n²-3n+3, We can't even factorise it, So let us , add 1 and Subtract 1 , 

=> n² - 3n + 3 + 1 - 1 ,
=> (n² -3n + 2) +1,
=> Going to Factorise the expression in the Brackets,
=> (n² - 1n -2n + 2) + 1,
=> (n(n-1) -2(n-1))+1,
=>((n-2)(n-1)) + 1,

Now, We need to check how many numbers are there , Which are perfect square when we keep value of n,

So, Assume The value is some k²,
=> (n-2)(n-1) + 1 = k²,
=> (n-2)(n-1) = k²-1²,
=> (n-2)(n-1) = (k+1)(k-1),  
=>  Let's use Equality method,
1st Case,
 Let n-2 = k -1,
=> n = k +1,
=> (k-1)(k) = k² - 1,
=> k² - k = k² - 1,
=> k =1,
=> n = 1+1 = 2,
2nd Case,
Let n-2 = k+1,
=> n = k +3,
=> (k+1)(k+2) = k²-1,
=> k² + 3k + 2 = k²-1,
=> 3k+ 2 = -1,
=> 3k = -3,
=> k = -1,
n = -1 + 3 = 2,
3rd Case
Let n-1 = k-1,
n = k,
=> (k-2)(k-1) = k² - 1,
=> k² -3k + 2 =k² -1,
=> -3k + 2 = -1,
=> -3k = -3,
=> k = 1,
=> n = 1,
4th Case,
Let n-1 = k+1,
n = k+2,
=> (k)(k+1) = k² -1,
=> k² + k = k² -1,
=> k = -1,
=> n = 1,
All the cases are over,
So by this , We can say that only n = 1,2 Satisfies the equation, 

Therefore the values of n which can make the given expression into a perfect square is (1,2) only, Also I have checked it by keeping every value of n from 1 to 20,

Hope you Understand, Have a Great day !
Thanking you,Bunti 360 !

And Sorry for the delay !