Prove that there is no three-digit number a b c ¯ such that a b c ¯ + b c a ¯ + c a b ¯ is a perfect square.

a b c ¯ + b c a ¯ + c a b ¯ = 100 a + 10 b + c + 100 b + 10 c + a + 100 c + 10 a + b = 111 a + b + c = 3 × 37 × a + b + c For a b c ¯ + b c a ¯ + c a b ¯ to be a perfect square, a + b + c must be equal to 3 × 37 = 111 . But the maximum value that a , b & c can take is 9 . So, 0 < a + b + c ≤ 9 + 9 + 9 = 27 . This contradicts that a + b + c = 111 . So, there is no three-digit number a b c ¯ such that a b c ¯ + b c a ¯ + c a b ¯ is a perfect square.

Prove that the number 3 n + 2 × 17 n , where n is a non-negative integer, is never a perfect square.

Let A = 3 n + 2 × 17 n , where n is a non-negative integer. We know that a perfect square number has remainder 0 , 1 , 4 , 5 , 6 or 9 modulo 10 . It is needed to consider several possible cases as follows : Case I : When n = 4 m , where m is a non-negative integer, then 3 n + 2 × 17 n = 3 4 m + 2 17 4 m = 81 m + 2 × 83521 m ≡ 3 mod 10 . So, A is not a perfect square. Case II : When n = 4 m + 1 , where m is a non-negative integer, then 3 n + 2 × 17 n = 3 × 81 m + 34 × 83521 m ≡ 7 mod 10 . So, A is not a perfect square. Case III : When n = 4 m + 2 , where m is a non-negative integer, then 3 n + 2 × 17 n = 9 × 81 m + 578 × 83521 m ≡ 7 mod 10 . So, A is not a perfect square. Case IV : When n = 4 m + 3 , where m is a non-negative integer, then 3 n + 2 × 17 n = 27 × 81 m + 9826 × 83521 m ≡ 3 mod 10 . So, A is not a perfect square. Thus, A is never a perfect square.