Find the Number $N$

Question

There is a number $N$ that if we add to him the number $100$ we take a square and if we add to him the number $168$ we take a square.Find this $N$

Any ideas would be apreciated for this puzzle

Answer 1

Let $s^2 = N+168$ and $t^2 = N+100$. Then $(s+t)(s-t) = s^2 - t^2 = 68 = 2\cdot 2 \cdot 17$. The choices for $s-t$ give $s=t+1, s=t+2, s=t+4, s=t+17, s=t+34, s=t+68$. (Restricting to positive integers because $\pm s$ and $\pm t$ work for any working choice of $s$ and $t$...)

• If $s = t+1$, then we started with $(t+1)^2 = N+168$ and subtracting $t^2 = N+100$, we are left with $2t+1 = 68$, which is impossible because the left side is odd and the right side is even.
• If $s = t+2$, then $(t+2)^2 = N+168$ and after subtraction, we are left with $4t+4 = 68$, which is satisfied by $t = 18$, so $s = 16$ and $N=156$.
• If $s = t+4$, then $(t+4)^2 = N+168$ and we are left with $8t+16 = 68$, which is unsatisfiable because $8$ does not divide $52$.
• If $s = t+17$, then $(t+17)^2 = N+168$ and then $34t+389 = 68$, which is unsatisfiable because $34$ does not divide $-321$. In fact this and the other choices cannot lead to positive $t$s, so we stop.

Having worked through the choices the only solution is $N = 156$.

Edit: Added a little more detail to the cases.

Answer 2

See that $N+100=A^2$ and $N+168=B^2$ for some $A$ and $B$, so $68=B^2-A^2=(B-A)(B+A).$ As $68=17\times 2^2$ it doesn't let a lot of choices for $B-A$ and $B+A,$ and by checking some cases you find $A=16, B=18$ and so $N=156.$