A number theory problem by Ossama Ismail

Number Theory Level 4

Let m m and n n are non-negative integers less than or equal to 1000. Determine the number of pairs of solution to n + m = ( n − m ) 2 . n+m = (n-m)^2 .


The answer is 89.

Ossama Ismail by Ossama Ismail , 63, Egypt

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1 solution

Kushal Bose
Mar 23, 2017

Consider n − m = k ⟹ n = m + k n-m=k \implies n=m+k

Now equation becomes 2 m + k = k 2 ⟹ m = k ( k − 1 ) 2 2m+k=k^2 \\ \implies m=\dfrac{k(k-1)}{2}

So, n = k + k ( k − 1 ) 2 = k ( k + 1 ) 2 n=k+\dfrac{k(k-1)}{2}=\dfrac{k(k+1)}{2}

As m , n ≤ 1000 m,n \leq 1000 So, k ( k + 1 ) 2 ≤ 1000 \dfrac{k(k+1)}{2} \leq 1000

A simple calculation gives k ≤ 44 k \leq 44

So, total number of ordered pairs 2 ∗ 45 − 1 = 89 2*45-1=89 because ( 0 , 0 ) (0,0) is counted twice.

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