SANSKAR'S OG PROBLEMS

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Problem 1

Let $\overline{ab}$ be a 2-digit positive integer satisfying $\overline{ab}^2=a! +b!$ . Find $a+b$ .

Solution 1 by ddk001 (Casework)

Case 1: $a>b$ In this case, we have

$\overline{ab}^2=a! +b!=(1+a \cdot (a-1) \cdot \dots \cdot (b+1)) \cdot b! \implies b!|\overline{ab}^2=(10a+b)^2$ .

If $b \ge 5$ , we must have

$10|b!|\overline{ab}^2=(10a+b)^2 \implies 10|(10a+b)^2=100a^2+20ab+b^2 \implies 10|b \implies b=0$

, but this contradicts the original assumption of $b \ge 5$ , so hence we must have $b \le 4$ .

Problem2

For any positive integer $n$ , $n$ >1 can $n!$ be a perfect square? If yes, give one such $n$ . If no, then prove it.

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SANSKAR'S OG PROBLEMS

Problem 1

Solution 1 by ddk001 (Casework)

Problem2