The sum of the squares of two consecutive natural numbers is 313 then numbers are

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Hint: Let the smaller number be $a$,then the second number will be $a+1$. Square these and add them and then equate the sum to 313. Then solve for $a$.Complete step-by-step answer:Let the smaller number be $a$. Now since the numbers are consecutive, the second number will be $a+1$.We are given that the sum of their squares is equal to 313.${{a}^{2}}+{{\left( a+1 \right)}^{2}}=313$ …(1)We know that ${{\left( x+y \right)}^{2}}={{x}^{2}}+{{y}^{2}}+2xy$. We will use this to simplify ${{\left( a+1 \right)}^{2}}$${{\left( a+1 \right)}^{2}}={{a}^{2}}+1+2a$We will substitute this in equation (1)${{a}^{2}}+{{a}^{2}}+1+2a=313$$2{{a}^{2}}+2a+1=313$$2{{a}^{2}}+2a-312=0$Divide this by 2, we will get the following:${{a}^{2}}+a-156=0$This is a quadratic equation in $a$.We know that $x=\dfrac{b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$where $a{{x}^{2}}+bx+c=0$ is a quadratic equation in $x$ .Using this, we get the following:$a=\dfrac{-1\pm \sqrt{{{\left( -1 \right)}^{2}}-4\times 1\times \left( -156 \right)}}{2\times 1}$ $a=\dfrac{-1\pm \sqrt{1+624}}{2}$$a=\dfrac{-1\pm \sqrt{625}}{2}$$a=\dfrac{-1\pm 25}{2}$$a=\dfrac{-26}{2},\dfrac{24}{2}$$a=-13,12$Since, in the question we need positive integers so we will select $a=12$So $a+1=13$Hence, the required consecutive positive integers are $12$ and $13$.Note: You can check whether your answer is correct or not by substituting $a=12$ and $a+1=13$ in the given equation ${{a}^{2}}+{{\left( a+1 \right)}^{2}}=313$. We get the following:${{12}^{2}}+{{13}^{2}}=144+169=313$. So our answer is correct.

The sum of the squares of two consecutive natural numbers is $$313$$. Then the numbers will be: