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Number theory Difficulty 6.2 National olympiad Prove it Iran

Let kZk \in \mathbb{Z} prove that there are infinitely many pairs of distinct positive integer numbers n,mn, m such that
n+S(2n)=m+S(2m),kn+S(n2)=km+S(m2), \begin{aligned} n + S(2n) &= m + S(2m), \\ kn + S(n^2) &= km + S(m^2), \end{aligned}
where S(n)S(n) is the sum of the digits of nn to base 10.

Solution

Let PkP_k be the set of solutions of
{n+S(n)=m+S(m),kn+S(kn)=km+S(km). \begin{cases} n + S(n) = m + S(m), \\ kn + S(kn) = km + S(km). \end{cases}
We want to map a single solution (m0,n0)Pk(m_0, n_0) \in P_k, where 10m0,n010 \nmid m_0, n_0, to infinite solutions like (m1,n1)(m_1, n_1), where (m1,n1)Pk+1(m_1, n_1) \in P_{k+1} or (m1,n1)Pk1(m_1, n_1) \in P_{k-1} and 10m1,n110 \nmid m_1, n_1.
If (m0,n0)(m_0, n_0) be a solution in PkP_k, then we claim that (m0+10α,n0+10α)(m_0 + 10^\alpha, n_0 + 10^\alpha) is a solution in Pk+1P_{k+1} for every large α\alpha. It follows by the same easy calculations and noting that
S(n2+2×10αn+102α)=S(n2)+S(2n)+1, S(n^2 + 2 \times 10^\alpha n + 10^{2\alpha}) = S(n^2) + S(2n) + 1,
for all large α\alpha.
And if (m0,n0)(m_0, n_0) is a solution in PkP_k where 10m0,n010 \nmid m_0, n_0, then we replace them by (10αm0,10αn0)(10^\alpha - m_0, 10^\alpha - n_0) for all large α\alpha to reach an infinite number of solutions in Pk1P_{k-1}. Therefore, it's enough to find a simple solution. For example we have (9,12)P0(9, 12) \in P_0 and we're done. ■

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