Question 182365
Let the consecutive numbers be n and (n+1) respectively.
Their sum is (n + n + 1) = (2n + 1)
Four times their sum = 4(2n + 1)
The product of the numbers = n(n + 1)
The product of two consecutive integers is 4 less than four times their sum.
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So, the equation becomes
n(n + 1) = 4 (2n + 1) - 4
n^2 + n = 8n + 4 - 4
n^2 + n = 8n
Subtract 8n on both the sides
n^2 + n - 8n = 8n - 8n
n^2 - 7n = 0
n (n - 7) = 0
Use the zero product rule
n=0 or (n-7)=0
n=0 or n=7.
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If n=0, n+1 = 0 + 1 = 1.
So, one pair of consecutive integers is 0 and 1.
If n=7, n+1 = 7 + 1 = 8.
So, second pair of consecutive integers is 7 and 8. 
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Cheers,
Mastermath.