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Question: Find two consecutive positive integers, the sum of whose squares is 365....

Find two consecutive positive integers, the sum of whose squares is 365.

Explanation

Solution

Consecutive positive integers: Consecutive positive integers are the positive integers that follow each other in order. They have a difference of 1 between every two positive integers.
If we take the first positive integer is mm then the second positive integers are m+1m + 1, third positive integers are m+2m + 2, and so on.
Formula used: (a+b)2=a2+2ab+b2{(a + b)^2} = {a^2} + 2ab + {b^2}

Complete step by step answer:
It is given that the sum of squares of two consecutive positive integers is 365.
Let, the first positive integer is m then the second positive integers is m+1.
Now the square of the first positive integer is m2{m^2}
The square of the second positive integer is (m+1)2{(m + 1)^2}
The sum of squares of two consecutive positive integers is m2+(m+1)2{m^2} + {(m + 1)^2}
Thus, we have
m2+(m+1)2=365{m^2} + {(m + 1)^2} = 365
Using the formula (a+b)2=a2+2ab+b2{(a + b)^2} = {a^2} + 2ab + {b^2}, we get,
m2+m2+2m+1=365{m^2} + {m^2} + 2m + 1 = 365
Simplifying as the equation, we get,
2m2+2m+1−365=02{m^2} + 2m + 1 - 365 = 0
2m2+2m−364=02{m^2} + 2m - 364 = 0
Solving the quadratic equation by middle term process
2m2+28m−26m−364=02{m^2} + 28m - 26m - 364 = 0
The above equation split into two terms,
2m(m+14)−26(m+14)=02m(m + 14) - 26(m + 14) = 0
Taking the common terms as a root we get,
(m+14)(2m−26)=0(m + 14)(2m - 26) = 0
Therefore the roots are,
m+14=0 or (2m−26)=0m + 14 = 0{\text{ or }}(2m - 26) = 0
Hence,
m=−14 or 262m = - 14{\text{ or }}\dfrac{{26}}{2}
m=−14 or 13m = - 14{\text{ or }}13
m = 13m{\text{ }} = {\text{ }}13
Since it is given that the consecutive numbers are positive integers.
So, the second positive integer is 13+1=1413 + 1 = 14
Hence, two consecutive positive integers are 13&1413\& 14.

Additional Information:
Whole numbers:
Whole numbers are simply the numbers 0, 1, 2, 3, 4, 5, 6…..0,{\text{ }}1,{\text{ }}2,{\text{ }}3,{\text{ }}4,{\text{ }}5,{\text{ }}6 \ldots ..

Integers:
Integers are like whole numbers, but they also include negative numbers but still, no fractions are allowed.

So integers can be negative \left\\{ { - 1,{\text{ }} - 2,{\text{ }} - 3,{\text{ }} \ldots \ldots } \right\\}, positive \left\\{ {1,{\text{ }}2,{\text{ }}3,{\text{ }}4,{\text{ }} \ldots \ldots } \right\\} or zero \left\\{ 0 \right\\}.

Note:
We have used the formula (a+b)2=a2+2ab+b2{(a + b)^2} = {a^2} + 2ab + {b^2}
A quadratic equation can be solved by the middle term process.
Splitting the middle term either addition of two numbers or subtraction of two numbers, then make the quadratic equation as the multiplication of two factors.
Solving the two factors we will get the solution.