# Three consecutive integers : correction of the exercises in third grade

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Three consecutive integers, corrected third grade math exercises on literal calculation.

Exercise:

a) Choose 3 consecutive integers.
Calculate the square of the middle number, then subtract the product of the other two numbers from this square.
b) Repeat with 3 more consecutive integers. What do we see?
c) Prove this conjecture.
Solution
a) Let be the three consecutive numbers 4 , 5 , 6 we have : 5² = 25 and 4×6 = 24 so 25-24=1
b) we proceed in the same way with the following triples:
7 ,8 ,9 ──► 8² = 64 and 7×9=63 so 64-63 = 1
10 , 11 , 12 ──► 11² = 121 and 10×12 = 120 so 121-120 = 1
147 , 148 ,149 ──► 148² = 21904 and 147×149 = 21903 so 21904-21903 = 1
We can see that the difference is always equal to 1
c) Let us prove this conjecture in the general case.
In order to simplify the calculations we choose the following triplet:
n-1 , n , n+1
(n-1)(n+1) = n²-1
and we have well
n² – (n²-1) = n² – n² + 1 = 1

Title:
Three consecutive integers
Correction:

Three consecutive integers, corrected third grade math exercises on literal calculation.

Type:

Level:
third

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